Solution\[ S_{x}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \qquad S_{y}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} \qquad S_{z}=\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \]
The first step is to find the eigenvalues of \(S_{z}\). These are the possible values that can be obtained when measuring the spin along the \(z\) axis. Because \(S_{z}\) is a diagonal matrix, its eigenvalues are on the diagonal. Hence the eigenvalues are \(\omega _{1}=0,\omega _{2}=1,\omega _{3}=-1\). Because the eigenvalues are different, \(S_{z}\) is not degenerate. The values are\begin {align*} \omega _{1} & =0\\ \omega _{2} & =1\\ \omega _{3} & =-1 \end {align*}
Now we need to find the eigenvalues for \(S_{y}\) and \(S_{x}\). The factor \(\frac {1}{\sqrt {2}}\) is not included in the following calculation, but added again at the end. This is to simplify the algebra.
For \(S_{y}\)\begin {align*} \left \vert S_{y}-\omega I\right \vert & =0\\\begin {vmatrix} -\omega & -i & 0\\ i & -\omega & -i\\ 0 & i & -\omega \end {vmatrix} & =0\\ -\omega \begin {vmatrix} -\omega & -i\\ i & -\omega \end {vmatrix} +i\begin {vmatrix} i & -i\\ 0 & -\omega \end {vmatrix} & =0\\ \left (-\omega \right ) \left (\omega ^{2}+i^{2}\right ) +i\left (-\omega i\right ) & =0\\ \left (-\omega \right ) \left (\omega ^{2}-1\right ) -\omega i^{2} & =0\\ -\omega ^{3}+\omega +\omega & =0\\ -\omega ^{3}+2\omega & =0\\ \omega \left (-\omega ^{2}+2\right ) & =0 \end {align*}
The eigenvalues are the roots of the above polynomial. They are\begin {align*} \omega _{1} & =0\\ \omega _{2} & =\sqrt {2}\\ \omega _{3} & =-\sqrt {2} \end {align*}
Adding back the factor \(\frac {1}{\sqrt {2}}\) which was in front of \(S_{y}\) by multiplying the above results with it gives\begin {align*} \omega _{1} & =0\\ \omega _{2} & =1\\ \omega _{3} & =-1 \end {align*}
For \(S_{x}\)\begin {align*} \left \vert S_{x}-\omega I\right \vert & =0\\\begin {vmatrix} -\omega & 1 & 0\\ 1 & -\omega & 1\\ 0 & 1 & -\omega \end {vmatrix} & =0\\ -\omega \begin {vmatrix} -\omega & 1\\ 1 & -\omega \end {vmatrix} -1\begin {vmatrix} 1 & 1\\ 0 & -\omega \end {vmatrix} & =0\\ \left (-\omega \right ) \left (\omega ^{2}-1\right ) -1\left (-\omega \right ) & =0\\ -\omega ^{3}+\omega +\omega & =0\\ -\omega ^{3}+2\omega & =0\\ \omega \left (2-\omega ^{2}\right ) & =0 \end {align*}
The eigenvalues are the roots of the above polynomial. They are\begin {align*} \omega _{1} & =0\\ \omega _{2} & =\sqrt {2}\\ \omega _{3} & =-\sqrt {2} \end {align*}
Adding back the factor \(\frac {1}{\sqrt {2}}\) which was in front of \(S_{y}\) by multiplying the above results with it gives\begin {align*} \omega _{1} & =0\\ \omega _{2} & =1\\ \omega _{3} & =-1 \end {align*}
| Spin matrix | Eigenvalues found |
| \(S_{x}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \) | \(\omega _{1}=0,\omega _{2}=1,\omega _{3}=-1\) |
| \(S_{y}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} \) | \(\omega _{1}=0,\omega _{2}=1,\omega _{3}=-1\) |
| \(S_{z}=\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \) | \(\omega _{1}=0,\omega _{2}=1,\omega _{3}=-1\) |
The above table shows that the possible values if we measure the spin along the \(x\) or \(y\) axis are \(\left \{ 0,1,-1\right \} \).
From part (b) and taking the largest eigenvalue of \(S_{x}\) as \(\omega _{2}=+1\), the question is asking us to find the associated eigenvector \(|S_{x}=\omega _{2}\rangle \). This is found by solving\[\begin {bmatrix} -\omega _{2} & \frac {1}{\sqrt {2}} & 0\\ \frac {1}{\sqrt {2}} & -\omega _{2} & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & -\omega _{2}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] In the above \(\omega _{2}=1\). Therefore\begin {equation} \begin {bmatrix} -1 & \frac {1}{\sqrt {2}} & 0\\ \frac {1}{\sqrt {2}} & -1 & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & -1 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \tag {1} \end {equation} \(R_{2}=R_{2}+\frac {1}{\sqrt {2}}R_{1}\) gives\[\begin {bmatrix} -1 & \frac {1}{\sqrt {2}} & 0\\ 0 & -\frac {1}{2} & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & -1 \end {bmatrix} \] \(R_{3}=R_{3}+\frac {2}{\sqrt {2}}R_{2}\)\[\begin {bmatrix} -1 & \frac {1}{\sqrt {2}} & 0\\ 0 & -\frac {1}{2} & \frac {1}{\sqrt {2}}\\ 0 & 0 & 0 \end {bmatrix} \] The above is now in Echelon form. The system becomes\[\begin {bmatrix} -1 & \frac {1}{\sqrt {2}} & 0\\ 0 & -\frac {1}{2} & \frac {1}{\sqrt {2}}\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] \(v_{3}\) is a free variable, and \(v_{2},v_{1}\) are the leading variables. Let \(v_{3}=s\). Second row gives \(-\frac {1}{2}v_{2}+\frac {1}{\sqrt {2}}s=0\) or \(v_{2}=\frac {2}{\sqrt {2}}s\). First row gives \(-v_{1}+\frac {1}{\sqrt {2}}v_{2}=0\) or \(v_{1}=\frac {1}{\sqrt {2}}v_{2}\) or \(v_{1}=\frac {1}{\sqrt {2}}\left (\frac {2}{\sqrt {2}}s\right ) =s\). Hence the solution (the eigenvector) is\begin {align*} \begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} s\\ \frac {2}{\sqrt {2}}s\\ s \end {bmatrix} \\ & =s\begin {bmatrix} 1\\ \frac {2}{\sqrt {2}}\\ 1 \end {bmatrix} \end {align*}
Since \(s\) is a free variable, we will choose it so that the norm is \(1\). Therefore\begin {align*} s\sqrt {1+2+1} & =1\\ s\sqrt {4} & =1\\ s & =\frac {1}{2} \end {align*}
Hence the state vector for the largest value of \(S_{x}\) is \[ |S_{x}=1\rangle =\frac {1}{2}\begin {bmatrix} 1\\ \frac {2}{\sqrt {2}}\\ 1 \end {bmatrix} =\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \]
\[ S_{z}=\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \] We first need to find the eigenvectors \(|S_{z}=\omega _{i}\rangle \) for \(S_{z}\). From part (a), the eigenvalues are\begin {align*} \omega _{1} & =0\\ \omega _{2} & =1\\ \omega _{3} & =-1 \end {align*}
For \(\omega _{1}=0\) the associated eigenvector is found by solving\begin {align*} \begin {bmatrix} 1-\omega _{1} & 0 & 0\\ 0 & -\omega _{1} & 0\\ 0 & 0 & -1-\omega _{1}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \end {align*}
\(v_{2}\) is a free variable, and \(v_{1},v_{3}\) are the leading variables. Let \(v_{2}=s\). Last row gives \(v_{3}=0\). First row gives \(v_{1}=0\). Hence the solution is\begin {align*} \begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ s\\ 0 \end {bmatrix} \\ & =s\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \end {align*}
Choosing \(s=1\) gives\[ |S_{z}=\omega _{1}\rangle =\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \] For \(\omega _{2}=1\) we need to solve\begin {align*} \begin {bmatrix} 1-\omega _{2} & 0 & 0\\ 0 & -\omega _{2} & 0\\ 0 & 0 & -1-\omega _{2}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 1-1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1-1 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 0 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -2 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \end {align*}
\(v_{1}\) is a free variable, and \(v_{2},v_{3}\) are the leading variables. Let \(v_{1}=s\). Last row gives \(v_{3}=0\). Second row gives \(v_{2}=0\). Hence the solution is\begin {align*} \begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} s\\ 0\\ 0 \end {bmatrix} \\ & =s\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \end {align*}
Choosing \(s=1\) then \[ |S_{z}=\omega _{2}\rangle =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \] For \(\omega _{3}=-1\) the associated eigenvector is found by solving\begin {align*} \begin {bmatrix} 1-\omega _{3} & 0 & 0\\ 0 & -\omega _{3} & 0\\ 0 & 0 & -1-\omega _{3}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 1+1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1+1 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \end {align*}
\(v_{3}\) is a free variable, and \(v_{1},v_{2}\) are the leading variables. Let \(v_{3}=s\). Second row gives \(v_{2}=0\). First row gives \(v_{1}=0\). Hence the solution is\begin {align*} \begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ s \end {bmatrix} \\ & =s\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \end {align*}
Choosing \(s=1\) gives \[ |S_{z}=\omega _{3}\rangle =\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \]
| eigenvalue | eigenvector |
| \(\omega _{1}=0\) | \(|S_{z}=0\rangle =\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \) |
| \(\omega _{2}=1\) | \(|S_{z}=1\rangle =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \) |
| \(\omega _{3}=-1\) | \(|S_{z}=-1\rangle =\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \) |
Calculating \(\left \vert \langle S_{z}=\omega _{1}|\Psi \rangle \right \vert ^{2}\) gives the odds of \(|S_{z}=\omega _{1}\rangle \). \(\Psi \) is the initial state vector. Similarly, calculating \(\left \vert \langle S_{x}=\omega _{1}|\Psi \rangle \right \vert ^{2}\) gives find the odds of \(|S_{x}=\omega _{1}\rangle \) and similarly for \(|S_{z}=\omega _{3}\rangle \).
\(\Psi \) is the state vector from part (c), which is\[ |\Psi \rangle =|S_{x}=1\rangle =\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \] Hence the odds of \(|S_{z}=0\rangle \) is\[ \left \vert \langle S_{z}=\omega _{1}|\Psi \rangle \right \vert ^{2}=\left ( \begin {bmatrix} 0 & 1 & 0 \end {bmatrix} ^{\ast }\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \right ) ^{2}=\left (\frac {1}{\sqrt {2}}\right ) ^{2}=\frac {1}{2}\] And the odds for \(|S_{z}=1\rangle \) is\[ \left \vert \langle S_{z}=\omega _{2}|\Psi \rangle \right \vert ^{2}=\left ( \begin {bmatrix} 1 & 0 & 0 \end {bmatrix} ^{\ast }\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \right ) ^{2}=\left (\frac {1}{2}\right ) ^{2}=\frac {1}{4}\] And the odds for \(|S_{z}=-1\rangle \) is\[ \left \vert \langle S_{z}=\omega _{3}|\Psi \rangle \right \vert ^{2}=\left ( \begin {bmatrix} 0 & 0 & 1 \end {bmatrix} ^{\ast }\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \right ) ^{2}=\left (\frac {1}{2}\right ) ^{2}=\frac {1}{4}\] The odds for \(|S_{z}=0\rangle \) is \(50\%,\) the odds for \(|S_{z}=1\rangle \) is \(25\%\) and odds for \(|S_{z}=-1\rangle \) is \(25\%\). The total is \(100\%\) as expected.
| eigenvalue | eigenvector | probability of this outcome |
| \(\omega _{1}=0\) | \(|S_{z}=\omega _{1}\rangle =\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \) | \(P\relax (0) =50\%\) |
| \(\omega _{2}=1\) | \(|S_{z}=\omega _{2}\rangle =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \) | \(P\relax (1) =25\%\) |
| \(\omega _{3}=-1\) | \(|S_{z}=\omega _{3}\rangle =\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \) | \(P\left (-1\right ) =25\%\) |
The state just after the measurement is \(|S_{z}=1\rangle =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \) since that is the state associated with the largest eigenvalue \(\omega _{2}=1\). This now becomes the initial state\[ |\Psi \rangle =|S_{z}=1\rangle =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \] We know that \(S_{x}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \) with the eigenvalues found earlier as \(\omega _{1}=0,\omega _{2}=1,\omega _{3}=-1\). In part (c) we found that \(|S_{x}=1\rangle =\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \) for \(S_{x}\) associated with its largest eigenvalue which is \(\omega _{2}=1\). Therefore the odds of this is\[ \left \vert \langle S_{x}=1|\Psi \rangle \right \vert ^{2}=\left ( \begin {bmatrix} \frac {1}{2} & \frac {1}{\sqrt {2}} & \frac {1}{2}\end {bmatrix} ^{\ast }\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \right ) ^{2}=\left (\frac {1}{2}\right ) ^{2}=\frac {1}{4}=25\% \] This says the odds of getting again the largest value (which is \(1\)) is not likely since it is not the highest possible odd being only \(25\%\) with \(3\) possible values.
\begin {align} S^{2} & =S_{x}^{2}+S_{y}^{2}+S_{z}^{3}\nonumber \\ & =\frac {1}{2}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} ^{2}+\frac {1}{2}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} ^{2}+\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} ^{2} \tag {1} \end {align}
But \begin {align*} \begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} ^{2} & =\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} =\begin {bmatrix} 1 & 0 & 1\\ 0 & 2 & 0\\ 1 & 0 & 1 \end {bmatrix} \\\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} ^{2} & =\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} =\begin {bmatrix} 1 & 0 & -1\\ 0 & 2 & 0\\ -1 & 0 & 1 \end {bmatrix} \\\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} ^{2} & =\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix}\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} =\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} \end {align*}
Hence (1) becomes\begin {align*} S^{2} & =\frac {1}{2}\begin {bmatrix} 1 & 0 & 1\\ 0 & 2 & 0\\ 1 & 0 & 1 \end {bmatrix} +\frac {1}{2}\begin {bmatrix} 1 & 0 & -1\\ 0 & 2 & 0\\ -1 & 0 & 1 \end {bmatrix} +\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} \\ & =\begin {bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 1 \end {bmatrix} +\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} \\ & =\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} \end {align*}
Since \(S^{2}\) is diagonal, then its eigenvalues are on the diagonal. They are all \(\omega =2\) with multiplicity \(3\). It is a degenerate matrix. Since the outcome is the eigenvalue (it is a measure of the spin angular momentum), then we see that the outcome is always \(2\), since that is the only possible eigenvalue.
The operators are\[ S^{2}=\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} \qquad S_{x}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \qquad S_{y}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} \qquad S_{z}=\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \] Commutator is defined as\[ \left [ M,N\right ] =MN-NM \] If \(\left [ M,N\right ] =0\) then they commute. We know that \(S_{x},S_{y},S_{z}\) do not commute with each others per lecture notes. So we only need to check if \(S^{2}\) commutes with \(S_{x},S_{y},S_{z}\) or not.\begin {align*} \left [ S^{2},S_{x}\right ] & =S^{2}S_{x}-S_{x}S^{2}\\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} -\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} \\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 2 & 0\\ 2 & 0 & 2\\ 0 & 2 & 0 \end {bmatrix} -\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 2 & 0\\ 2 & 0 & 2\\ 0 & 2 & 0 \end {bmatrix} \\ & =\begin {bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end {bmatrix} \end {align*}
Hence \(S^{2},S_{x}\) commute. And
\begin {align*} \left [ S^{2},S_{y}\right ] & =S^{2}S_{y}-S_{y}S^{2}\\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} -\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} \\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & -2i & 0\\ 2i & 0 & -2i\\ 0 & 2i & 0 \end {bmatrix} -\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & -2i & 0\\ 2i & 0 & -2i\\ 0 & 2i & 0 \end {bmatrix} \\ & =\begin {bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end {bmatrix} \end {align*}
Hence \(S^{2},S_{y}\) commute. And\begin {align*} \left [ S^{2},S_{z}\right ] & =S^{2}S_{z}-S_{z}S^{2}\\ & =\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} \\ & =\begin {bmatrix} 2 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -2 \end {bmatrix} -\begin {bmatrix} 2 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -2 \end {bmatrix} \\ & =\begin {bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end {bmatrix} \end {align*}
Hence \(S^{2},S_{z}\) commute. Therefore there are three sets of commuting operators. They are \(\left \{ S^{2},S_{x}\right \} ,\left \{ S^{2},S_{y}\right \} ,\left \{ S^{2},S_{z}\right \} \). So the maximum number of operators such that they all commute with each others is two.
\[ |V\rangle =\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} \] The norm is \(\sqrt {1+4+9}=\sqrt {14}\). Hence the normalized state is\[ |V\rangle =\frac {1}{\sqrt {14}}\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} \] In part (c) we found the eigenvalues and associated eigenvector for \(S_{z}\). Here they are again
| eigenvalue | eigenvector |
| \(\omega _{1}=0\) | \(|S_{z}=\omega _{1}\rangle =\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \) |
| \(\omega _{2}=1\) | \(|S_{z}=\omega _{2}\rangle =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \) |
| \(\omega _{3}=-1\) | \(|S_{z}=\omega _{3}\rangle =\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \) |
We will now find the odds of getting \(|S_{z}=\omega _{1}\rangle \) given the current state vector is \(|V\rangle \) (after normalizing). The odds are \[ \left \vert \langle S_{z}=0|V\rangle \right \vert ^{2}=\left (\frac {1}{\sqrt {14}}\begin {bmatrix} 0 & 1 & 0 \end {bmatrix}\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} \right ) ^{2}=\left (\frac {2}{\sqrt {14}}\right ) ^{2}=\frac {4}{14}=28.571\% \] And\[ \left \vert \langle S_{z}=+1|V\rangle \right \vert ^{2}=\left (\frac {1}{\sqrt {14}}\begin {bmatrix} 1 & 0 & 0 \end {bmatrix}\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} \right ) ^{2}=\left (\frac {1}{\sqrt {14}}\right ) ^{2}=\frac {1}{14}=7.143\% \] And\[ \left \vert \langle S_{z}=-1|V\rangle \right \vert ^{2}=\left (\frac {1}{\sqrt {14}}\begin {bmatrix} 0 & 0 & 1 \end {bmatrix}\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} \right ) ^{2}=\left (\frac {3}{\sqrt {14}}\right ) ^{2}=\frac {9}{14}=64.285\% \]
| eigenvalue | eigenvector | odd of getting this eigenvalue |
| \(\omega _{1}=0\) | \(|S_{z}=\omega _{1}\rangle =\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \) | \(P\relax (0) =\frac {4}{14}=28.571\%\) |
| \(\omega _{2}=1\) | \(|S_{z}=\omega _{2}\rangle =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \) | \(P\relax (1) =\frac {1}{14}=7.143\%\) |
| \(\omega _{3}=-1\) | \(|S_{z}=\omega _{3}\rangle =\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \) | \(P\left (-1\right ) =\frac {9}{14}=64.285\%\) |
The statistical average is \begin {align} \omega _{1}\left (\frac {4}{14}\right ) +\omega _{2}\left (\frac {1}{14}\right ) +\omega _{3}\left (\frac {9}{14}\right ) & =0\left (\frac {4}{14}\right ) +1\left (\frac {1}{14}\right ) -1\left (\frac {9}{14}\right ) \nonumber \\ & =-\frac {4}{7}\nonumber \\ & =-0.571\,43 \tag {1} \end {align}
The above is now compared to Now we compare \(\langle V|S_{z}|V\rangle \) \begin {equation} \langle V|\left (S_{z}|V\right ) \rangle =\frac {1}{\sqrt {14}}\langle V|\overset {S_{z}|V\rangle }{\overbrace {\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix}\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} }}\rangle \tag {2} \end {equation} But \[ S_{z}|V\rangle =\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix}\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} =\begin {bmatrix} 1\\ 0\\ -3 \end {bmatrix} \] Hence Eq. (2) becomes\begin {align} \langle V|\left (S_{z}|V\right ) \rangle & =\frac {1}{\sqrt {14}}\left ( \frac {1}{\sqrt {14}}\begin {bmatrix} 1 & 2 & 3 \end {bmatrix} ^{\ast }\begin {bmatrix} 1\\ 0\\ -3 \end {bmatrix} \right ) \nonumber \\ & =\frac {1}{14}\left (1-9\right ) \nonumber \\ & =\frac {-8}{14}\nonumber \\ & =-0.571\,43 \tag {3} \end {align}
Comparing (1) and (3) shows it is the same value. This is the expectation value when measuring \(S_{z}\).
Part (g) is now repeated, but using \(S_{x}\). We found from the above part that \[ |V\rangle =\frac {1}{\sqrt {14}}\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} \] From part(b), we found the eigenvalues for \(S_{x}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \) to be \(\omega _{1}=0,\omega _{2}=1,\omega _{3}=-1\). But we did not find the associated eigenvectors yet in order to repeat part g as was done for \(S_{z}\). So we need now to find the eigenvectors for \(S_{x}\) before being able to answer this part for \(S_{x}\).
For \(\omega _{1}=0\)
\[\begin {bmatrix} 0 & \frac {1}{\sqrt {2}} & 0\\ \frac {1}{\sqrt {2}} & 0 & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] Swapping \(R_{2},R_{1}\)\[\begin {bmatrix} \frac {1}{\sqrt {2}} & 0 & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & 0\\ 0 & \frac {1}{\sqrt {2}} & 0 \end {bmatrix} \] \(R_{3}=R_{3}-R_{2}\)\[\begin {bmatrix} \frac {1}{\sqrt {2}} & 0 & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & 0\\ 0 & 0 & 0 \end {bmatrix} \] Now it is in echelon form. Hence the system becomes\[\begin {bmatrix} \frac {1}{\sqrt {2}} & 0 & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & 0\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] \(v_{3}\) is free variable. Let \(v_{3}=s\). Second row gives \(v_{2}=0\). First row gives \(\frac {1}{\sqrt {2}}v_{1}+\frac {1}{\sqrt {2}}s=0\) or \(v_{1}=-s\). Hence solution is\[\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} -s\\ 0\\ s \end {bmatrix} =s\begin {bmatrix} -1\\ 0\\ 1 \end {bmatrix} \] Let \(s=\frac {1}{\sqrt {2}}\). Therefore\[ |S_{x}=\omega _{1}\rangle =\frac {1}{\sqrt {2}}\begin {bmatrix} -1\\ 0\\ 1 \end {bmatrix} =\begin {bmatrix} \frac {-1}{\sqrt {2}}\\ 0\\ \frac {1}{\sqrt {2}}\end {bmatrix} \] For \(\omega _{2}=1\)\begin {align*} \begin {bmatrix} -\omega _{2} & \frac {1}{\sqrt {2}} & 0\\ \frac {1}{\sqrt {2}} & -\omega _{2} & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & -\omega _{2}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} -1 & \frac {1}{\sqrt {2}} & 0\\ \frac {1}{\sqrt {2}} & -1 & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & -1 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \end {align*}
\(R_{2}=R_{2}+\frac {1}{\sqrt {2}}R_{1}\)\[\begin {bmatrix} -1 & \frac {1}{\sqrt {2}} & 0\\ 0 & -\frac {1}{2} & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & -1 \end {bmatrix} \] \(R_{3}=R_{3}+\frac {2}{\sqrt {2}}R_{2}\)\[\begin {bmatrix} -1 & \frac {1}{\sqrt {2}} & 0\\ 0 & -\frac {1}{2} & \frac {1}{\sqrt {2}}\\ 0 & 0 & 0 \end {bmatrix} \] Now it is in echelon form. Hence the system becomes\[\begin {bmatrix} -1 & \frac {1}{\sqrt {2}} & 0\\ 0 & -\frac {1}{2} & \frac {1}{\sqrt {2}}\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] \(v_{3}\) is free variable. Let \(v_{3}=s\). Second row gives \(-\frac {1}{2}v_{2}+\frac {1}{\sqrt {2}}s=0\) or \(v_{2}=\frac {2}{\sqrt {2}}s\). First row gives \(-v_{1}+\frac {1}{\sqrt {2}}v_{2}=0\) or \(v_{1}=\frac {1}{\sqrt {2}}\left ( \frac {2}{\sqrt {2}}s\right ) =s\). Hence solution is\[\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} s\\ \frac {2}{\sqrt {2}}s\\ s \end {bmatrix} =s\begin {bmatrix} 1\\ \frac {2}{\sqrt {2}}\\ 1 \end {bmatrix} =s\begin {bmatrix} 1\\ \sqrt {2}\\ 1 \end {bmatrix} \] Let \(s=\frac {1}{2}\). Therefore\[ |S_{x}=\omega _{2}\rangle =\frac {1}{2}\begin {bmatrix} 1\\ \sqrt {2}\\ 1 \end {bmatrix} =\begin {bmatrix} \frac {1}{2}\\ \frac {\sqrt {2}}{2}\\ \frac {1}{2}\end {bmatrix} =\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \] For \(\omega _{3}=-1\)\begin {align*} \begin {bmatrix} -\omega _{3} & \frac {1}{\sqrt {2}} & 0\\ \frac {1}{\sqrt {2}} & -\omega _{3} & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & -\omega _{3}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} 1 & \frac {1}{\sqrt {2}} & 0\\ \frac {1}{\sqrt {2}} & 1 & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & 1 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \end {align*}
\(R_{2}=R_{2}-\frac {1}{\sqrt {2}}R_{1}\)\[\begin {bmatrix} 1 & \frac {1}{\sqrt {2}} & 0\\ 0 & \frac {1}{2} & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & 1 \end {bmatrix} \] \(R_{3}=R_{3}-\frac {2}{\sqrt {2}}R_{2}\)\[\begin {bmatrix} 1 & \frac {1}{\sqrt {2}} & 0\\ 0 & \frac {1}{2} & \frac {1}{\sqrt {2}}\\ 0 & 0 & 0 \end {bmatrix} \] Now it is in echelon form. Hence the system becomes\[\begin {bmatrix} 1 & \frac {1}{\sqrt {2}} & 0\\ 0 & \frac {1}{2} & \frac {1}{\sqrt {2}}\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \] \(v_{3}\) is free variable. Let \(v_{3}=s\). Second row gives \(\frac {1}{2}v_{2}+\frac {1}{\sqrt {2}}s=0\) or \(v_{2}=-\frac {2}{\sqrt {2}}s\). First row gives \(v_{1}+\frac {1}{\sqrt {2}}v_{2}=0\) or \(v_{1}=-\frac {1}{\sqrt {2}}\left ( -\frac {2}{\sqrt {2}}s\right ) =s\). Hence solution is\[\begin {bmatrix} v_{1}\\ v_{2}\\ v_{3}\end {bmatrix} =\begin {bmatrix} s\\ -\frac {2}{\sqrt {2}}s\\ s \end {bmatrix} =s\begin {bmatrix} 1\\ -\frac {2}{\sqrt {2}}\\ 1 \end {bmatrix} =s\begin {bmatrix} 1\\ -\sqrt {2}\\ 1 \end {bmatrix} \] Let \(s=\frac {1}{2}\). Therefore\[ |S_{x}=\omega _{3}\rangle =\frac {1}{2}\begin {bmatrix} 1\\ -\sqrt {2}\\ 1 \end {bmatrix} =\begin {bmatrix} \frac {1}{2}\\ -\frac {\sqrt {2}}{2}\\ \frac {1}{2}\end {bmatrix} =\begin {bmatrix} \frac {1}{2}\\ -\frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \]
| eigenvalue | eigenvector |
| \(\omega _{1}=0\) | \(|S_{x}=\omega _{1}\rangle =\begin {bmatrix} \frac {-1}{\sqrt {2}}\\ 0\\ \frac {1}{\sqrt {2}}\end {bmatrix} \) |
| \(\omega _{2}=1\) | \(|S_{x}=\omega _{2}\rangle =\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \) |
| \(\omega _{3}=-1\) | \(|S_{x}=\omega _{3}\rangle =\begin {bmatrix} \frac {1}{2}\\ -\frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \) |
The odds of getting \(|S_{x}=\omega _{i}\rangle \) given the current state vector is \(|V\rangle \) are now found. Expressing \(|V\rangle \) in the eigenbasis of \(S_{x}\) gives\begin {align} |V\rangle & =c_{1}\ |S_{x}=\omega _{1}\rangle +c_{2}\ |S_{x}=\omega _{2}\rangle +c_{3}\ |S_{x}=\omega _{3}\rangle \nonumber \\ & =c_{1}\ |S_{x}=0\rangle +c_{2}\ |S_{x}=1\rangle +c_{3}\ |S_{x}=-1\rangle \tag {1} \end {align}
Where\begin {align*} c_{1} & =\langle S_{x}=0|V\rangle =\frac {1}{\sqrt {14}}\begin {bmatrix} \frac {-1}{\sqrt {2}} & 0 & \frac {1}{\sqrt {2}}\end {bmatrix} ^{\ast }\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} =\frac {1}{\sqrt {14}}\left (\frac {-1}{\sqrt {2}}+\frac {3}{\sqrt {2}}\right ) =\frac {1}{\sqrt {14}}\left (\frac {2}{\sqrt {2}}\right ) \\ c_{2} & =\langle S_{x}=1|V\rangle =\frac {1}{\sqrt {14}}\begin {bmatrix} \frac {1}{2} & \frac {1}{\sqrt {2}} & \frac {1}{2}\end {bmatrix} ^{\ast }\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} =\frac {1}{\sqrt {14}}\left (\frac {1}{2}+\frac {2}{\sqrt {2}}+\frac {3}{2}\right ) =\frac {1}{\sqrt {14}}\left (\frac {2}{\sqrt {2}}\right ) \\ c_{3} & =\langle S_{x}=-1|V\rangle =\frac {1}{\sqrt {14}}\begin {bmatrix} \frac {1}{2} & -\frac {1}{\sqrt {2}} & \frac {1}{2}\end {bmatrix} ^{\ast }\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} =\frac {1}{\sqrt {14}}\left (\frac {1}{2}-\frac {2}{\sqrt {2}}+\frac {3}{2}\right ) =\frac {1}{\sqrt {14}}\left (2-\sqrt {2}\right ) \end {align*}
Eq. (1) becomes\[ |V\rangle =\frac {1}{\sqrt {14}}\left (\frac {2}{\sqrt {2}}\right ) \begin {bmatrix} \frac {-1}{\sqrt {2}}\\ 0\\ \frac {1}{\sqrt {2}}\end {bmatrix} +\frac {1}{\sqrt {14}}\left (\sqrt {2}+2\right ) \begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} +\frac {1}{\sqrt {14}}\left (2-\sqrt {2}\right ) \begin {bmatrix} \frac {1}{2}\\ -\frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \] The above is the representation of \(|V\rangle \) in the eigenbasis of \(S_{x}\). The odds of each eigenvalue is the square of the coefficients \(\left \vert c_{1}\right \vert ^{2},\left \vert c_{2}\right \vert ^{2},\left \vert c_{3}\right \vert ^{2}\) above. Therefore\begin {align*} P\relax (0) & =\left (\frac {1}{\sqrt {14}}\left (\frac {2}{\sqrt {2}}\right ) \right ) ^{2}=\frac {2}{14}=14.286\%\\ P\left (+1\right ) & =\left (\frac {1}{\sqrt {14}}\left (\sqrt {2}+2\right ) \right ) ^{2}=\frac {1}{14}\left (6+4\sqrt {2}\right ) =83.263\%\\ P\left (-1\right ) & =\left (\frac {1}{\sqrt {14}}\left (2-\sqrt {2}\right ) \right ) ^{2}=\frac {1}{14}\left (6-4\sqrt {2}\right ) =24.51\% \end {align*}
| eigenvalue | eigenvector | Odds of getting this eigenvalue |
| \(\omega _{1}=0\) | \(|S_{x}=\omega _{1}\rangle =\begin {bmatrix} \frac {-1}{\sqrt {2}}\\ 0\\ \frac {1}{\sqrt {2}}\end {bmatrix} \) | \(P\relax (0) =\frac {2}{14}=14.286\%\) |
| \(\omega _{2}=1\) | \(|S_{x}=\omega _{2}\rangle =\begin {bmatrix} \frac {1}{2}\\ \frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \) | \(P\relax (1) =\frac {1}{14}\left (6+4\sqrt {2}\right ) =83.263\%\) |
| \(\omega _{3}=-1\) | \(|S_{x}=\omega _{3}\rangle =\begin {bmatrix} \frac {1}{2}\\ -\frac {1}{\sqrt {2}}\\ \frac {1}{2}\end {bmatrix} \) | \(P\left (-1\right ) =\frac {1}{14}\left (6-4\sqrt {2}\right ) =24.51\%\) |
The statistical average is \begin {align} \omega _{1}\left (\frac {2}{14}\right ) +\omega _{2}\left (\frac {1}{14}\left ( 6+4\sqrt {2}\right ) \right ) +\omega _{3}\left (\frac {1}{14}\left ( 6-4\sqrt {2}\right ) \right ) & =0\left (\frac {2}{14}\right ) +1\left ( \frac {1}{14}\left (6+4\sqrt {2}\right ) \right ) -1\left (\frac {1}{14}\left ( 6-4\sqrt {2}\right ) \right ) \nonumber \\ & =\frac {4}{7}\sqrt {2}\nonumber \\ & =0.80812 \tag {1} \end {align}
The above is now compared to\begin {equation} \langle V|S_{x}|V\rangle =\frac {1}{\sqrt {14}}\langle V|\overset {S_{x}|V\rangle }{\overbrace {\begin {bmatrix} 0 & \frac {1}{\sqrt {2}} & 0\\ \frac {1}{\sqrt {2}} & 0 & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & 0 \end {bmatrix}\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} }}\rangle \tag {2} \end {equation} But \[ S_{x}|V\rangle =\begin {bmatrix} 0 & \frac {1}{\sqrt {2}} & 0\\ \frac {1}{\sqrt {2}} & 0 & \frac {1}{\sqrt {2}}\\ 0 & \frac {1}{\sqrt {2}} & 0 \end {bmatrix}\begin {bmatrix} 1\\ 2\\ 3 \end {bmatrix} =\begin {bmatrix} \sqrt {2}\\ 2\sqrt {2}\\ \sqrt {2}\end {bmatrix} \] Hence Eq. (2) becomes\begin {align} \langle V|S_{x}|V\rangle & =\frac {1}{\sqrt {14}}\left (\frac {1}{\sqrt {14}}\begin {bmatrix} 1 & 2 & 3 \end {bmatrix} ^{\ast }\begin {bmatrix} \sqrt {2}\\ 2\sqrt {2}\\ \sqrt {2}\end {bmatrix} \right ) \nonumber \\ & =\frac {1}{14}\left (\sqrt {2}+4\sqrt {2}+3\sqrt {2}\right ) \nonumber \\ & =0.80812 \tag {3} \end {align}
Comparing (1) and (3) shows it is the same value.This is the expectation value when measuring \(S_{x}\)
Prove the following results on commutators: \begin {align*} \left [ A,B+C\right ] & =\left [ A,B\right ] +\left [ A,C\right ] \\ \left [ A+B,C\right ] & =\left [ A,C\right ] +\left [ B,C\right ] \\ \left [ A,BC\right ] & =B\left [ A,C\right ] +\left [ A,B\right ] C\\ \left [ AB,C\right ] & =A\left [ B,C\right ] +\left [ A,C\right ] B \end {align*}
Solution
By definition of commutator, which is \(\left [ A,B\right ] =AB-BA\), then\begin {align*} \left [ A,B+C\right ] & =A\left (B+C\right ) -\left (B+C\right ) A\\ & =AB+AC-BA-CA\\ & =\left (AB-BA\right ) +\left (AC-CA\right ) \\ & =\left [ A,B\right ] +\left [ A,C\right ] \end {align*}
By definition of commutator, which is \(\left [ A,B\right ] =AB-BA\), then\begin {align*} \left [ A+B,C\right ] & =\left (A+B\right ) C-C\left (A+B\right ) \\ & =AC+BC-CA-CB\\ & =\left (AC-CA\right ) +\left (BC-CB\right ) \\ & =\left [ A,C\right ] +\left [ B,C\right ] \end {align*}
By definition of commutator, which is \(\left [ A,B\right ] =AB-BA\), then\[ \left [ A,BC\right ] =A\left (BC\right ) -\left (BC\right ) A \] Adding and subtracting \(BAC\) on the RHS gives\begin {align*} \left [ A,BC\right ] & =\widehat {BAC}+ABC-BCA-\widehat {BAC}\\ & =\left (BAC-BCA\right ) +\left (ABC-BAC\right ) \\ & =B\left (AC-CA\right ) +\left (AB-BA\right ) C\\ & =B\left [ A,C\right ] +\left [ A,B\right ] C \end {align*}
By definition of commutator, which is \(\left [ A,B\right ] =AB-BA\), then\begin {align*} \left [ AB,C\right ] & =\left (AB\right ) C-C\left (AB\right ) \\ & =ABC-CAB \end {align*}
Adding and subtracting \(ACB\) on the RHS gives\begin {align*} \left [ AB,C\right ] & =\widehat {ACB}+ABC-CAB-\widehat {ACB}\\ & =\left (ABC-ACB\right ) +\left (ACB-CAB\right ) \\ & =A\left (BC-CB\right ) +\left (AC-CA\right ) B\\ & =A\left [ B,C\right ] +\left [ A,C\right ] B \end {align*}
Follow the discussion of \(s_{+}=s_{x}+is_{y}\) for the electron spin to derive the matrix representation of \(s_{-}=s_{x}-is_{y}\)
Solution
Experiments show that \(S_{z}\) has two possible values (eigenvalues) of \(\frac {\hslash }{2},-\frac {\hslash }{2}\). Using eigenbasis of \(S_{z}\)\begin {align*} |S_{z} & =\frac {\hslash }{2}\rangle =\begin {bmatrix} 1\\ 0 \end {bmatrix} =|1\rangle \\ |S_{z} & =-\frac {\hslash }{2}\rangle =\begin {bmatrix} 0\\ 1 \end {bmatrix} =|2\rangle \end {align*}
Gives\[ S_{z}=\frac {\hslash }{2}\begin {bmatrix} 1 & 0\\ 0 & -1 \end {bmatrix} \] Consider \(S_{-}=S_{x}-iS_{y}\). Then\begin {align} \left [ S_{z},S_{-}\right ] & =\left [ S_{z},S_{x}-iS_{y}\right ] \nonumber \\ & =\left [ S_{z},S_{x}\right ] -i\left [ S_{z},S_{y}\right ] \tag {1} \end {align}
But, using \(\left [ S_{i},S_{j}\right ] =i\hslash \sum _{k}\epsilon _{ijk}S_{k}\). Hence \begin {align} \left [ S_{z},S_{x}\right ] & =i\hslash S_{y}\tag {2}\\ \left [ S_{z},S_{y}\right ] & =-i\hslash S_{x} \tag {3} \end {align}
Substituting (2,3) into (1) gives\begin {align*} \left [ S_{z},S_{-}\right ] & =i\hslash S_{y}-i\left (-i\hslash S_{x}\right ) \\ & =i\hslash S_{y}+i^{2}\left (\hslash S_{x}\right ) \\ & =i\hslash S_{y}-\hslash S_{x}\\ & =\hslash \left (iS_{y}-S_{x}\right ) \\ & =-\hslash \left (S_{x}-iS_{y}\right ) \\ & =-\hslash S_{-} \end {align*}
Therefore we see that\[ \left [ S_{z},S_{-}\right ] =S_{z}S_{-}-S_{-}S_{z}=-\hslash S_{-}\] This implies\[ S_{z}S_{-}=S_{-}S_{z}-\hslash S_{-}\] Therefore\begin {align*} S_{z}S_{-}|1\rangle & =\left (S_{-}S_{z}-\hslash S_{-}\right ) |1\rangle \\ & =S_{-}S_{z}|1\rangle -\hslash S_{-}|1\rangle \end {align*}
But \(S_{z}|1\rangle =\frac {\hslash }{2}|1\rangle \) then the above becomes\begin {align*} S_{z}S_{-}|1\rangle & =S_{-}\frac {\hslash }{2}|1\rangle -\hslash S_{-}|1\rangle \\ & =\left (\frac {\hslash }{2}-\hslash \right ) S_{-}|1\rangle \\ & =-\frac {\hslash }{2}S_{-}|1\rangle \end {align*}
The above shows that \(S_{-}|1\rangle \) is eigenvector (eigenstate) of \(S_{z}\) with eigenvalue \(-\frac {\hslash }{2}\) which is compatible with experiments. Because \(S_{z}|2\rangle =-\frac {\hslash }{2}|2\rangle \) then let\begin {equation} S_{-}|1\rangle =c|2\rangle \tag {4} \end {equation} We now need to find \(c\). Taking the adjoint of both sides of (4) gives\[ \langle 1|S_{-}^{\dag }=c^{\ast }\langle 2| \] Therefore\begin {align*} \langle 1|S_{-}^{\dag }\ S_{-}|1\rangle & =c^{\ast }c\langle 2|2\rangle \\ & =\left \vert c\right \vert ^{2}\langle 2|2\rangle \end {align*}
Since \(c\) is real. But \(\langle 2|2\rangle =\begin {bmatrix} 0 & 1 \end {bmatrix}\begin {bmatrix} 0\\ 1 \end {bmatrix} =1\). The above becomes\begin {equation} \langle 1|S_{-}^{\dag }\ S_{-}|1\rangle =\left \vert c\right \vert ^{2} \tag {5} \end {equation} To find \(c\), we need now to calculate \(\langle 1|S_{-}^{\dag }\ S_{-}|1\rangle \). But \begin {align*} S_{-}^{\dag }\ S_{-} & =\left (S_{x}-iS_{y}\right ) ^{\dag }\left ( S_{x}-iS_{y}\right ) \\ & =\left (S_{x}^{\dag }+iS_{y}^{\dag }\right ) \left (S_{x}-iS_{y}\right ) \end {align*}
Since \(S_{x},S_{y}\) are Hermitian operators then \(S_{x}^{\dag }=S_{x}\) and \(S_{y}^{\dag }=S_{y}\). The above now becomes\begin {align*} S_{-}^{\dag }\ S_{-} & =\left (S_{x}+iS_{y}\right ) \left (S_{x}-iS_{y}\right ) \\ & =S_{x}^{2}-iS_{x}S_{y}+iS_{y}S_{x}+S_{y}^{2}\\ & =S_{x}^{2}+S_{y}^{2}-i\left (S_{x}S_{y}-S_{y}S_{x}\right ) \\ & =S_{x}^{2}+S_{y}^{2}-i[S_{x},S_{y}] \end {align*}
Where \([S_{x},S_{y}]\) is the commutator. But \([S_{x},S_{y}]=i\hbar \sum _{k}\epsilon _{ijk}S_{k}\). Using \(i=1,j=2\) for \(x,y\), then \([S_{i},S_{i}]=i\hbar \left (\epsilon _{121}S_{1}+\epsilon _{122}S_{2}+\epsilon _{123}S_{3}\right ) =i\hbar S_{3}=i\hbar S_{z}\). Therefore the above now becomes\begin {align} S_{-}^{\dag }\ S_{-} & =S_{x}^{2}+S_{y}^{2}-i\left (i\hbar S_{z}\right ) \nonumber \\ & =S_{x}^{2}+S_{y}^{2}+\hbar S_{z} \tag {6} \end {align}
Substituting (6) in (5) gives\[ \langle 1|\left (S_{x}^{2}+S_{y}^{2}+\hbar S_{z}\right ) |1\rangle =\left \vert c\right \vert ^{2}\] But \(S^{2}=S_{x}^{2}+S_{y}^{2}+S_{z}^{2}\). Hence \(S_{x}^{2}+S_{y}^{2}=S^{2}-S_{z}^{2}\). Using this in the above gives\begin {equation} \langle 1|\left (S^{2}-S_{z}^{2}+\hbar S_{z}\right ) |1\rangle =\left \vert c\right \vert ^{2} \tag {7} \end {equation} But \(S_{z}=\frac {\hbar }{2}\begin {bmatrix} 1 & 0\\ 0 & -1 \end {bmatrix} \). Hence \(S_{z}^{2}=\frac {\hbar ^{2}}{4}\begin {bmatrix} 1 & 0\\ 0 & -1 \end {bmatrix}\begin {bmatrix} 1 & 0\\ 0 & -1 \end {bmatrix} =\frac {\hbar ^{2}}{4}\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix} =\frac {\hbar ^{2}}{4}I\). And since there is nothing special about the \(z\) direction, then \(S_{x}^{2}=S_{y}^{2}=S_{z}^{2}\). Therefore \(S^{2}=S_{x}^{2}+S_{y}^{2}+S_{z}^{2}=\frac {\hbar ^{2}}{4}I+\frac {\hbar ^{2}}{4}I+\frac {\hbar ^{2}}{4}I=\frac {3}{4}\hbar ^{2}I\). Eq. (7) now becomes\[ \langle 1|\frac {3}{4}\hbar ^{2}-\frac {\hbar ^{2}}{4}+\hbar S_{z}|1\rangle =\left \vert c\right \vert ^{2}\] But \(S_{z}|1\rangle =\frac {\hbar }{2}|1\rangle \). This is because \(|1\rangle \) is an eigenvector for \(S_{z}\) with an eigenvalue \(\frac {\hbar }{2}\). The above becomes\begin {align*} \langle 1|\frac {3}{4}\hbar ^{2}-\frac {\hbar ^{2}}{4}+\hbar \frac {\hbar }{2}|1\rangle & =\left \vert c\right \vert ^{2}\\ \langle 1|\frac {3}{4}\hbar ^{2}-\frac {\hbar ^{2}}{4}+\frac {\hbar ^{2}}{2}|1\rangle & =\left \vert c\right \vert ^{2}\\ \langle 1|\hbar ^{2}|1\rangle & =\left \vert c\right \vert ^{2}\\ \hbar ^{2}\langle 1|1\rangle & =\left \vert c\right \vert ^{2}\\ \hbar ^{2} & =\left \vert c\right \vert ^{2} \end {align*}
We pick \[ c=\hbar \] Now that \(c\) is found, then Eq. (4) above becomes\begin {equation} S_{-}|1\rangle =\hbar |2\rangle \tag {8} \end {equation} The same method is now repeated for finding \(S_{-}|2\rangle \)\begin {align*} S_{z}S_{-}|2\rangle & =\left (S_{-}S_{z}-\hslash S_{-}\right ) |2\rangle \\ & =S_{-}S_{z}|2\rangle -\hslash S_{-}|2\rangle \end {align*}
But \(S_{z}|2\rangle =-\frac {\hslash }{2}|2\rangle \). The above becomes\begin {align*} S_{z}S_{-}|2\rangle & =-S_{-}\frac {\hslash }{2}|2\rangle -\hslash S_{-}|2\rangle \\ & =\left (-\frac {\hslash }{2}-\hslash \right ) S_{-}|2\rangle \\ & =\left (-\frac {3\hslash }{2}\right ) S_{-}|2\rangle \end {align*}
The above shows that \(S_{-}|2\rangle \) is eigenvector (eigenstate) of \(S_{z}\) with eigenvalue \(-\frac {3\hslash }{2}\) which conflicts with experiments. This means\begin {equation} S_{-}|2\rangle =0|2\rangle \tag {9} \end {equation} is the only logical result. Therefore now we have all the information to find matrix representation of\(\ S\_\) using (8,9), which is \begin {align*} S_{-} & =\begin {bmatrix} \langle 1|S_{-}|1\rangle & \langle 1|S_{-}|2\rangle \\ \langle 2|S_{-}|1\rangle & \langle 2|S_{-}|2\rangle \end {bmatrix} \\ & =\begin {bmatrix} \langle 1|\hslash |2\rangle & \langle 1|0|2\rangle \\ \langle 2|\hslash |2\rangle & \langle 2|0|2\rangle \end {bmatrix} \\ & =\hslash \begin {bmatrix} \langle 1|2\rangle & 0\\ \langle 2|2\rangle & 0 \end {bmatrix} \\ & =\hslash \begin {bmatrix}\begin {bmatrix} 1 & 0 \end {bmatrix}\begin {bmatrix} 0\\ 1 \end {bmatrix} & 0\\\begin {bmatrix} 0 & 1 \end {bmatrix}\begin {bmatrix} 0\\ 1 \end {bmatrix} & 0 \end {bmatrix} \\ & =\hslash \begin {bmatrix} 0 & 0\\ 1 & 0 \end {bmatrix} \end {align*}
Therefore\begin {align*} S_{-} & =S_{x}-iS_{y}\\ & =\hslash \begin {bmatrix} 0 & 0\\ 1 & 0 \end {bmatrix} \end {align*}
Which is what we are asked to show.
Find the solutions \(x_{1}\relax (t) ,x_{2}\relax (t) \) with initial conditions \(x_{1}\relax (0) =0,x_{2}\relax (0) =0\) and \(\dot {x}_{1}\relax (0) =v_{1},\dot {x}_{2}\relax (0) =v_{2}.\)
Solution
The first step is to draw the free body diagram for each mass. Let us assume that first mass is at some positive distance \(x_{1}>0\) so that the first string is in tension, and that \(x_{2}>x_{1}>0\) so that the middle spring is in tension also, and the third spring is in compression. Any other configuration will also work as well. Based on this, the free body diagrams are
From the free body diagram, we can now write the equation of motion based on \(F=ma\) from each mass. This gives\begin {align*} m\ddot {x}_{1} & =-kx_{1}+2k\left (x_{2}-x_{1}\right ) \\ m\ddot {x}_{2} & =-2k\left (x_{2}-x_{1}\right ) -kx_{2} \end {align*}
or\begin {align*} m\ddot {x}_{1} & =-kx_{1}+2kx_{2}-2kx_{1}\\ m\ddot {x}_{2} & =-2kx_{2}+2kx_{1}-kx_{2} \end {align*}
or\begin {align*} m\ddot {x}_{1} & =x_{1}\left (-k-2k\right ) +x_{2}\left (2k\right ) \\ m\ddot {x}_{2} & =x_{1}\left (2k\right ) +x_{2}\left (-2k-k\right ) \end {align*}
or\begin {align*} \ddot {x}_{1} & =-\frac {3k}{m}x_{1}+\frac {2k}{m}x_{2}\\ \ddot {x}_{2} & =\frac {2k}{m}x_{1}-3\frac {k}{m}x_{2} \end {align*}
In matrix form the above becomes\begin {align} \begin {bmatrix} \ddot {x}_{1}\\ \ddot {x}_{2}\end {bmatrix} & =\frac {k}{m}\begin {bmatrix} -3 & 2\\ 2 & -3 \end {bmatrix}\begin {bmatrix} x_{1}\\ x_{2}\end {bmatrix} \nonumber \\\begin {bmatrix} \ddot {x}_{1}\\ \ddot {x}_{2}\end {bmatrix} +\frac {k}{m}\begin {bmatrix} 3 & -2\\ -2 & 3 \end {bmatrix}\begin {bmatrix} x_{1}\\ x_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \nonumber \\ |\ddot {x}\rangle +M|x\rangle & =|0\rangle \tag {1} \end {align}
Where the operator \(M\) is\begin {equation} M=\frac {k}{m}\begin {bmatrix} 3 & -2\\ -2 & 3 \end {bmatrix} \tag {1A} \end {equation} In (1), the state vector is \(|x\rangle \) is represented using basis \(|1\rangle =\begin {bmatrix} 1\\ 0 \end {bmatrix} \) and \(|2\rangle =\begin {bmatrix} 0\\ 1 \end {bmatrix} \), since we can write \[ |x\rangle =x_{1}|1\rangle +x_{2}|2\rangle \] In these basis, called the natural coordinates, we see than operator \(M\) is not diagonal. This makes solving (1) harder, since it is now a coupled system of ODE’s.
We would like to decouple (1) to make solving each ODE separate and easier. To do this, we change the basis of \(M\). The new basis are \(|I\rangle ,|II\rangle \). These are the eigenvectors of \(M\). Since \(M\) is Hermitian, then its eigenvalues will be real, and its eigenvectors are orthogonal. So now we need to first find the eigenvalues of \(M\) given in (1A) by solving
\begin {align*} \det \left (M-\omega I\right ) & =0\\ \det \left (\frac {k}{m}\begin {bmatrix} 3 & -2\\ -2 & 3 \end {bmatrix} -\omega \begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix} \right ) & =0\\ \det \left (\frac {k}{m}\begin {bmatrix} 3-\omega & -2\\ -2 & 3-\omega \end {bmatrix} \right ) & =0 \end {align*}
This gives (we remove the factor \(\frac {k}{m}\) for now, then add it at the end to simplify the computation)\begin {align*} \left (3-\omega \right ) \left (3-\omega \right ) -4 & =0\\ \omega ^{2}-6\omega +5 & =0\\ \left (\omega -5\right ) \left (\omega -1\right ) & =0 \end {align*}
Hence the eigenvalues are (now we add back the factor \(\frac {k}{m}\)) \[ \omega _{1}=\frac {5k}{m}\qquad \omega _{1}=\frac {k}{m}\] For \(\omega _{1}=\frac {k}{m}\)
We need to solve\begin {align*} \begin {bmatrix} \frac {3k}{m}-\omega _{1} & -\frac {2k}{m}\\ -\frac {2k}{m} & \frac {3k}{m}-\omega _{1}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \\\begin {bmatrix} \frac {3k}{m}-\frac {k}{m} & -\frac {2k}{m}\\ -\frac {2k}{m} & \frac {3k}{m}-\frac {k}{m}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \\\begin {bmatrix} \frac {2k}{m} & -\frac {2k}{m}\\ -\frac {2k}{m} & \frac {2k}{m}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \end {align*}
\(R_{2}=R_{2}+R_{1}\)\[\begin {bmatrix} \frac {2k}{m} & -\frac {2k}{m}\\ 0 & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \] Hence \(v_{2}\) is free variable. Let \(v_{2}=s\). First row gives \(\frac {2k}{m}v_{1}-\frac {2k}{m}s=0\) or \(v_{1}=s\). Hence solution is\[\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} =\begin {bmatrix} s\\ s \end {bmatrix} =s\begin {bmatrix} 1\\ 1 \end {bmatrix} \] Let \(s=\frac {1}{\sqrt {2}}\) then\[ |I\rangle =|M=\omega _{1}\rangle =\frac {1}{\sqrt {2}}\begin {bmatrix} 1\\ 1 \end {bmatrix} \] For \(\omega _{1}=5\frac {k}{m}\)
We need to solve\begin {align*} \begin {bmatrix} \frac {3k}{m}-\omega _{2} & -\frac {2k}{m}\\ -\frac {2k}{m} & \frac {3k}{m}-\omega _{2}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \\\begin {bmatrix} \frac {3k}{m}-5\frac {k}{m} & -\frac {2k}{m}\\ -\frac {2k}{m} & \frac {3k}{m}-5\frac {k}{m}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \\\begin {bmatrix} -\frac {2k}{m} & -\frac {2k}{m}\\ -\frac {2k}{m} & -\frac {2k}{m}\end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \end {align*}
\(R_{2}=R_{2}-R_{1}\)\[\begin {bmatrix} -\frac {2k}{m} & -\frac {2k}{m}\\ 0 & 0 \end {bmatrix}\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \] Hence \(v_{2}\) is free variable. Let \(v_{2}=s\). First row gives \(-\frac {2k}{m}v_{1}-\frac {2k}{m}s=0\) or \(v_{1}=-s\). Hence solution is\[\begin {bmatrix} v_{1}\\ v_{2}\end {bmatrix} =\begin {bmatrix} -s\\ s \end {bmatrix} =s\begin {bmatrix} -1\\ 1 \end {bmatrix} \] Let \(s=\frac {1}{\sqrt {2}}\) then\[ |II\rangle =|M=\omega _{2}\rangle =\frac {1}{\sqrt {2}}\begin {bmatrix} -1\\ 1 \end {bmatrix} \]
| eigenvalue | eigenfrequncy | eigenvector |
| \(\omega _{1}=\frac {k}{m}\) | \(\sqrt {\frac {k}{m}}\) | \(|I\rangle =\frac {1}{\sqrt {2}}\begin {bmatrix} 1\\ 1 \end {bmatrix} \) |
| \(\omega _{2}=5\frac {k}{m}\) | \(\sqrt {\frac {5k}{m}}\) | \(|II\rangle =\frac {1}{\sqrt {2}}\begin {bmatrix} -1\\ 1 \end {bmatrix} \) |
The transformation matrix \(\Phi \) becomes \begin {align} \Phi & =\begin {bmatrix} |I\rangle & |II\rangle \end {bmatrix} \nonumber \\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix} \tag {2} \end {align}
Now that we found the transformation matrix \(\Phi \) we can use it to transform \(|\ddot {x}\rangle +M|x\rangle =0\) which is the natural coordinates basis \(|1\rangle ,|2\rangle \), to the modal coordinates based on basis \(|I\rangle ,|II\rangle \) as follows\begin {align} |x\rangle & =\Phi |X\rangle \tag {3}\\ |\ddot {x}\rangle & =\Phi |\ddot {X}\rangle \tag {4} \end {align}
Where \[ |X\rangle =X_{1}|I\rangle +X_{2}|II\rangle \] is the state vector in the modal coordinate and \(|\ddot {X}\rangle \) is the acceleration of the state vector in modal coordinates. Applying Eq. (3,4) to \(|\ddot {x}\rangle =M|x\rangle \) gives the system in the modal coordinates as\[ \Phi |\ddot {X}\rangle +M\Phi |X\rangle =0 \] Premultiplying both sides by \(\Phi ^{T}\) (since \(\Phi \) is real, then transpose is same as dagger), gives\begin {equation} \Phi ^{T}\Phi |\ddot {X}\rangle +\Phi ^{T}M\Phi |X\rangle =0 \tag {5} \end {equation} But by definition of the modal transformation matrix1 \begin {equation} \Phi ^{T}\Phi =I \tag {6} \end {equation} And by definition of the transformation matrix 2 \begin {align} \Phi ^{T}M\Phi & =\begin {bmatrix} \omega _{1} & 0\\ 0 & \omega _{2}\end {bmatrix} \nonumber \\ & =\begin {bmatrix} \frac {k}{m} & 0\\ 0 & \frac {5k}{m}\end {bmatrix} \tag {7} \end {align}
Using (6,7) in (5) gives the system in modal coordinates\begin {align} |\ddot {X}\rangle +\begin {bmatrix} \frac {k}{m} & 0\\ 0 & \frac {5k}{m}\end {bmatrix} |X\rangle & =0\nonumber \\\begin {bmatrix} \ddot {X}_{1}\\ \ddot {X}_{2}\end {bmatrix} +\begin {bmatrix} \frac {k}{m} & 0\\ 0 & \frac {5k}{m}\end {bmatrix}\begin {bmatrix} X_{1}\\ X_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \tag {8} \end {align}
The above is what solve, since it is now decoupled. Comparing (8) to (1) which is repeated below\begin {equation} \begin {bmatrix} \ddot {x}_{1}\\ \ddot {x}_{2}\end {bmatrix} +\begin {bmatrix} \frac {3k}{m} & -\frac {2k}{m}\\ -\frac {2k}{m} & \frac {3k}{m}\end {bmatrix}\begin {bmatrix} x_{1}\\ x_{2}\end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \tag {1} \end {equation} Shows clearly why (8) is much simpler to solve in the modal coordinates basis \(|I\rangle ,|II\rangle \) since it is now decoupled, while Eq (1) which is in natural coordinates basis \(|1\rangle ,|2\rangle \) is coupled.
Eq (8) is now solved for \(|X\rangle \), and at the end transformed back to \(|x\rangle \) using Eq. (3). Eq (8) above can be written as two separate ODE’s\begin {align*} \ddot {X}_{1}+\frac {k}{m}X_{1} & =0\\ \ddot {X}_{2}+\frac {5k}{m}X_{2} & =0 \end {align*}
Before solving the above, the initial conditions, given in the natural coordinates, needs to be transformed to modal coordinates. It is not clear which initial conditions we should use, since book uses \begin {align*} x_{1}\left (t=0\right ) & =x_{1}\relax (0) \qquad \dot {x}_{1}\left ( t=0\right ) =0\\ x_{2}\left (t=0\right ) & =x_{2}\relax (0) \qquad \dot {x}_{2}\left ( t=0\right ) =0 \end {align*}
And in the HW pdf, we are also asked to use the following initial conditions\begin {align*} x_{1}\left (t=0\right ) & =0\qquad \dot {x}_{1}\left (t=0\right ) =v_{1}\\ x_{2}\left (t=0\right ) & =0\qquad \dot {x}_{2}\left (t=0\right ) =v_{2} \end {align*}
Should we solve it for both cases, or just the second case? I will solve the problem for both cases, since I am not sure which to pick.
Solving using book initial conditions \begin {align*} x_{1}\left (t=0\right ) & =x_{1}\relax (0) \qquad \dot {x}_{1}\left ( 0\right ) =0\\ x_{2}\left (t=0\right ) & =x_{2}\relax (0) \qquad \dot {x}_{2}\left ( 0\right ) =0 \end {align*}
Since \(|x\rangle =\Phi |X\rangle \) then the inverse is\[ |X\rangle =\Phi ^{-1}|x\rangle \] But \(\Phi ^{-1}=\Phi ^{T}\) therefore\begin {align*} |X\relax (0) \rangle & =\Phi ^{T}|x_{1}\relax (0) \rangle \\\begin {bmatrix} X_{1}\relax (0) \\ X_{2}\relax (0) \end {bmatrix} & =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix} ^{T}\begin {bmatrix} x_{1}\relax (0) \\ x_{2}\relax (0) \end {bmatrix} \\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & 1\\ -1 & 1 \end {bmatrix}\begin {bmatrix} x_{1}\relax (0) \\ x_{2}\relax (0) \end {bmatrix} \\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} x_{1}\relax (0) +x_{2}\relax (0) \\ -x_{1}\relax (0) +x_{2}\relax (0) \end {bmatrix} \end {align*}
And\begin {align*} |\dot {X}\relax (0) \rangle & =\Phi ^{T}|\dot {x}_{1}\relax (0) \rangle \\\begin {bmatrix} \dot {X}_{1}\relax (0) \\ \dot {X}_{2}\relax (0) \end {bmatrix} & =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & 1\\ -1 & 1 \end {bmatrix}\begin {bmatrix} \dot {x}_{1}\relax (0) \\ \dot {x}_{2}\relax (0) \end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \end {align*}
Since \(\begin {bmatrix} \dot {x}_{1}\relax (0) \\ \dot {x}_{2}\relax (0) \end {bmatrix} =\begin {bmatrix} 0\\ 0 \end {bmatrix} \). Now that we found the initial conditions in modal coordinates, we can solve Eq. (8). Here it is again\begin {align} \begin {bmatrix} \ddot {X}_{1}\\ \ddot {X}_{2}\end {bmatrix} +\begin {bmatrix} \frac {k}{m} & 0\\ 0 & \frac {5k}{m}\end {bmatrix}\begin {bmatrix} X_{1}\\ X_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \tag {9}\\\begin {bmatrix} X_{1}\relax (0) \\ X_{2}\relax (0) \end {bmatrix} & =\frac {1}{\sqrt {2}}\begin {bmatrix} x_{1}\relax (0) +x_{2}\relax (0) \\ -x_{1}\relax (0) +x_{2}\relax (0) \end {bmatrix} \nonumber \\\begin {bmatrix} \dot {X}_{1}\relax (0) \\ \dot {X}_{2}\relax (0) \end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \nonumber \end {align}
The first equation of (9) becomes\begin {align*} \ddot {X}_{1}+\frac {k}{m}X_{1} & =0\\ X_{1}\relax (0) & =\frac {1}{\sqrt {2}}\left (x_{1}\relax (0) +x_{2}\relax (0) \right ) \\ \dot {X}_{1}\relax (0) & =0 \end {align*}
The solution is\begin {equation} X_{1}\relax (t) =A\cos \left (\sqrt {\frac {k}{m}}t\right ) +B\sin \left ( \sqrt {\frac {k}{m}}t\right ) \tag {10} \end {equation} Where \(A,B\) are the constants of integrations. At \(t=0\) and from the initial conditions, the above becomes\[ \frac {1}{\sqrt {2}}\left (x_{1}\relax (0) +x_{2}\relax (0) \right ) =A \] Taking time derivative of (10) gives\[ \dot {X}_{1}=-A\sqrt {\frac {k}{m}}\sin \left (\sqrt {\frac {k}{m}}t\right ) +B\sqrt {\frac {k}{m}}\cos \left (\sqrt {\frac {k}{m}}t\right ) \] Since \(\dot {X}_{1}\relax (0) =0\) then the above becomes\[ 0=B\sqrt {\frac {k}{m}}\] Hence \(B=0\). Therefore the solution of Eq (10) is\begin {equation} X_{1}=\frac {1}{\sqrt {2}}\left (x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left (\sqrt {\frac {k}{m}}t\right ) \tag {11} \end {equation} Tthe second ODE in (9) is now solved.\begin {align*} \ddot {X}_{2}+\frac {5k}{m}X_{1} & =0\\ X_{2}\relax (0) & =\frac {1}{\sqrt {2}}\left (-x_{1}\relax (0) +x_{2}\relax (0) \right ) \\ \dot {X}_{2}\relax (0) & =0 \end {align*}
The solution is\begin {equation} X_{2}=A\cos \left (\sqrt {\frac {5k}{m}}t\right ) +B\sin \left (\sqrt {\frac {5k}{m}}t\right ) \tag {12} \end {equation} Where \(A,B\) are the constants of integrations. At \(t=0\), \[ \frac {1}{\sqrt {2}}\left (-x_{1}\relax (0) +x_{2}\relax (0) \right ) =A \] Taking time derivative of (12) gives\[ \dot {X}_{2}=-A\sqrt {\frac {5k}{m}}\sin \left (\sqrt {\frac {5k}{m}}t\right ) +B\sqrt {\frac {5k}{m}}\cos \left (\sqrt {\frac {5k}{m}}t\right ) \] At \(t=0\) the above becomes\[ 0=B\sqrt {\frac {5k}{m}}\] Hence \(B=0\). The solution of Eq (12) becomes\begin {equation} X_{2}=\frac {1}{\sqrt {2}}\left (-x_{1}\relax (0) +x_{2}\left ( 0\right ) \right ) \cos \left (\sqrt {\frac {5k}{m}}t\right ) \tag {13} \end {equation} Therefore the solution is\begin {equation} |X\rangle =\begin {bmatrix} \frac {1}{\sqrt {2}}\left (x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left (\sqrt {\frac {k}{m}}t\right ) \\ \frac {1}{\sqrt {2}}\left (-x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left (\sqrt {\frac {5k}{m}}t\right ) \end {bmatrix} \tag {14} \end {equation} This is the final solution. But it is in modal coordinates. This is transformed back to natural coordinates using Eq (3)\[ |x\rangle =\Phi |X\rangle \] Therefore\begin {align} |x\rangle & =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix}\begin {bmatrix} \frac {1}{\sqrt {2}}\left (x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left (\sqrt {\frac {k}{m}}t\right ) \\ \frac {1}{\sqrt {2}}\left (-x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left (\sqrt {\frac {5k}{m}}t\right ) \end {bmatrix} \nonumber \\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} \frac {1}{\sqrt {2}}\left (x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left (\sqrt {\frac {k}{m}}t\right ) -\frac {1}{\sqrt {2}}\left ( -x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left ( \sqrt {\frac {5k}{m}}t\right ) \\ \frac {1}{\sqrt {2}}\left (x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left (\sqrt {\frac {k}{m}}t\right ) +\frac {1}{\sqrt {2}}\left ( -x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left ( \sqrt {\frac {5k}{m}}t\right ) \end {bmatrix} \nonumber \\ & =\frac {1}{2}\begin {bmatrix} \left (x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left ( \sqrt {\frac {k}{m}}t\right ) +\left (x_{1}\relax (0) -x_{2}\left ( 0\right ) \right ) \cos \left (\sqrt {\frac {5k}{m}}t\right ) \\ \left (x_{1}\relax (0) +x_{2}\relax (0) \right ) \cos \left ( \sqrt {\frac {k}{m}}t\right ) -\left (x_{1}\relax (0) -x_{2}\left ( 0\right ) \right ) \cos \left (\sqrt {\frac {5k}{m}}t\right ) \end {bmatrix} \tag {15} \end {align}
Hence\begin {align} x_{1}\relax (t) & =\frac {x_{1}\relax (0) +x_{2}\left ( 0\right ) }{2}\cos \left (\sqrt {\frac {k}{m}}t\right ) +\frac {x_{1}\left ( 0\right ) -x_{2}\relax (0) }{2}\cos \left (\sqrt {\frac {5k}{m}}t\right ) \tag {16}\\ x_{2}\relax (t) & =\frac {x_{1}\relax (0) +x_{2}\left ( 0\right ) }{2}\cos \left (\sqrt {\frac {k}{m}}t\right ) -\frac {x_{1}\left ( 0\right ) -x_{2}\relax (0) }{2}\cos \left (\sqrt {\frac {5k}{m}}t\right ) \tag {17} \end {align}
The above is the final solution in the natural coordinates. The above is repeated using the other initial conditions given in the PDF file.
Solving using book initial conditions \begin {align*} x_{1}\left (t=0\right ) & =0\qquad \dot {x}_{1}\left (t=0\right ) =v_{1}\\ x_{2}\left (t=0\right ) & =0\qquad \dot {x}_{2}\left (t=0\right ) =v_{2} \end {align*}
Using \(|x\rangle =\Phi |X\rangle \) then\[ |X\rangle =\Phi ^{-1}|x\rangle \] But \(\Phi ^{-1}=\Phi ^{T}\) then \begin {align*} |X\relax (0) \rangle & =\Phi ^{T}|x_{1}\relax (0) \rangle \\\begin {bmatrix} X_{1}\relax (0) \\ X_{2}\relax (0) \end {bmatrix} & =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix} ^{T}\begin {bmatrix} x_{1}\relax (0) \\ x_{2}\relax (0) \end {bmatrix} \\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & 1\\ -1 & 1 \end {bmatrix}\begin {bmatrix} 0\\ 0 \end {bmatrix} \\ & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \end {align*}
And\begin {align*} |\dot {X}\relax (0) \rangle & =\Phi ^{T}|\dot {x}_{1}\relax (0) \rangle \\\begin {bmatrix} \dot {X}_{1}\relax (0) \\ \dot {X}_{2}\relax (0) \end {bmatrix} & =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & 1\\ -1 & 1 \end {bmatrix}\begin {bmatrix} \dot {x}_{1}\relax (0) \\ \dot {x}_{2}\relax (0) \end {bmatrix} \\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} v_{1}+v_{2}\\ -v_{1}+v_{2}\end {bmatrix} \end {align*}
Now that we found initial conditions in modal coordinates, we can finally solve the (8). Here it is again\begin {align} \begin {bmatrix} \ddot {X}_{1}\\ \ddot {X}_{2}\end {bmatrix} +\begin {bmatrix} \frac {k}{m} & 0\\ 0 & \frac {5k}{m}\end {bmatrix}\begin {bmatrix} X_{1}\\ X_{2}\end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \tag {18}\\\begin {bmatrix} X_{1}\relax (0) \\ X_{2}\relax (0) \end {bmatrix} & =\begin {bmatrix} 0\\ 0 \end {bmatrix} \nonumber \\\begin {bmatrix} \dot {X}_{1}\relax (0) \\ \dot {X}_{2}\relax (0) \end {bmatrix} & =\frac {1}{\sqrt {2}}\begin {bmatrix} v_{1}+v_{2}\\ -v_{1}+v_{2}\end {bmatrix} \nonumber \end {align}
The first equation of (18) becomes\begin {align*} \ddot {X}_{1}+\frac {k}{m}X_{1} & =0\\ X_{1}\relax (0) & =0\\ \dot {X}_{1}\relax (0) & =\frac {1}{\sqrt {2}}\left (v_{1}+v_{2}\right ) \end {align*}
The solution is (since SHM) \begin {equation} X_{1}=A\cos \left (\sqrt {\frac {k}{m}}t\right ) +B\sin \left (\sqrt {\frac {k}{m}}t\right ) \tag {19} \end {equation} Where \(A,B\) are the constants of integrations. At \(t=0\), \[ 0=A \] The solution (19) becomes\[ X_{1}=B\sin \left (\sqrt {\frac {k}{m}}t\right ) \] Taking time derivative of the above gives\[ \dot {X}_{1}=B\sqrt {\frac {k}{m}}\cos \left (\sqrt {\frac {k}{m}}t\right ) \] At \(t=0\) the above becomes\begin {align*} \frac {1}{\sqrt {2}}\left (v_{1}+v_{2}\right ) & =B\sqrt {\frac {k}{m}}\\ B & =\sqrt {\frac {m}{k}}\frac {1}{\sqrt {2}}\left (v_{1}+v_{2}\right ) \end {align*}
Therefore the solution of Eq (19) is\begin {equation} X_{1}=\sqrt {\frac {m}{2k}}\left (v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {k}{m}}t\right ) \tag {20} \end {equation} The second ODE in (18) is now solved.\begin {align*} \ddot {X}_{2}+\frac {5k}{m}X_{1} & =0\\ X_{2}\relax (0) & =0\\ \dot {X}_{2}\relax (0) & =\frac {1}{\sqrt {2}}\left (-v_{1}+v_{2}\right ) \end {align*}
The solution is\begin {equation} X_{2}=A\cos \left (\sqrt {\frac {5k}{m}}t\right ) +B\sin \left (\sqrt {\frac {5k}{m}}t\right ) \tag {21} \end {equation} Where \(A,B\) are the constants of integrations. At \(t=0\), \[ 0=A \] The solution (21) becomes\[ X_{2}=B\sin \left (\sqrt {\frac {5k}{m}}t\right ) \] Taking time derivative gives\[ \dot {X}_{2}=B\sqrt {\frac {5k}{m}}\cos \left (\sqrt {\frac {5k}{m}}t\right ) \] At \(t=0\) the above becomes\begin {align*} \frac {1}{\sqrt {2}}\left (-v_{1}+v_{2}\right ) & =B\sqrt {\frac {5k}{m}}\\ B & =\sqrt {\frac {m}{10k}}\left (-v_{1}+v_{2}\right ) \end {align*}
Therefore the solution of Eq (21) is\begin {equation} X_{2}=\sqrt {\frac {m}{10k}}\left (-v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {5k}{m}}t\right ) \tag {22} \end {equation} Therefore the solution state vector is \begin {equation} |X\rangle =\begin {bmatrix} \sqrt {\frac {m}{2k}}\left (v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {k}{m}}t\right ) \\ \sqrt {\frac {m}{10k}}\left (-v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {5k}{m}}t\right ) \end {bmatrix} \tag {23} \end {equation} This is the final solution. But it is in modal coordinates. It is now transformed back to natural coordinates using Eq (3)\[ |x\rangle =\Phi |X\rangle \] Therefore\begin {align} |x\rangle & =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix}\begin {bmatrix} \sqrt {\frac {m}{2k}}\left (v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {k}{m}}t\right ) \\ \sqrt {\frac {m}{10k}}\left (-v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {5k}{m}}t\right ) \end {bmatrix} \nonumber \\ & =\frac {1}{\sqrt {2}}\begin {bmatrix} \sqrt {\frac {m}{2k}}\left (v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {k}{m}}t\right ) -\sqrt {\frac {m}{10k}}\left (-v_{1}+v_{2}\right ) \sin \left ( \sqrt {\frac {5k}{m}}t\right ) \\ \sqrt {\frac {m}{2k}}\left (v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {k}{m}}t\right ) +\sqrt {\frac {m}{10k}}\left (-v_{1}+v_{2}\right ) \sin \left ( \sqrt {\frac {5k}{m}}t\right ) \end {bmatrix} \nonumber \\ & =\begin {bmatrix} \sqrt {\frac {m}{4k}}\left (v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {k}{m}}t\right ) -\sqrt {\frac {m}{20k}}\left (-v_{1}+v_{2}\right ) \sin \left ( \sqrt {\frac {5k}{m}}t\right ) \\ \sqrt {\frac {m}{4k}}\left (v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {k}{m}}t\right ) +\sqrt {\frac {m}{20k}}\left (-v_{1}+v_{2}\right ) \sin \left ( \sqrt {\frac {5k}{m}}t\right ) \end {bmatrix} \tag {24} \end {align}
Hence\begin {align} x_{1}\relax (t) & =\frac {1}{2}\sqrt {\frac {m}{k}}\left (v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {k}{m}}t\right ) +\sqrt {\frac {m}{20k}}\left (v_{1}-v_{2}\right ) \sin \left (\sqrt {\frac {5k}{m}}t\right ) \tag {25}\\ x_{2}\relax (t) & =\frac {1}{2}\sqrt {\frac {m}{k}}\left (v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {k}{m}}t\right ) +\sqrt {\frac {m}{20k}}\left (-v_{1}+v_{2}\right ) \sin \left (\sqrt {\frac {5k}{m}}t\right ) \tag {26} \end {align}
The above is the final solution in the natural coordinates.
1This can also be shown for \(\Phi =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix} \) by working it out. \(\Phi ^{T}\Phi =\frac {1}{2}\begin {bmatrix} 1 & 1\\ -1 & 1 \end {bmatrix}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix} =\frac {1}{2}\begin {bmatrix} 2 & 0\\ 0 & 2 \end {bmatrix} =\begin {bmatrix} 1 & 0\\ 0 & 1 \end {bmatrix} \)
2This can also be shown by working it out as follows. \(\Phi ^{T}M\Phi =\frac {1}{\sqrt {2}}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix} ^{T}\frac {k}{m}\begin {bmatrix} 3 & -2\\ -2 & 3 \end {bmatrix} \frac {1}{\sqrt {2}}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix} \) which becomes \(\Phi ^{T}M\Phi =\frac {k}{2m}\begin {bmatrix} 1 & 1\\ -1 & 1 \end {bmatrix}\begin {bmatrix} 3 & -2\\ -2 & 3 \end {bmatrix}\begin {bmatrix} 1 & -1\\ 1 & 1 \end {bmatrix} =\allowbreak \begin {bmatrix} \frac {k}{m} & 0\\ 0 & \frac {5k}{m}\end {bmatrix} \)