Problem says that the system is in some general state \(\psi \relax (x) \) and asks what is the probability distribution to measure momentum \(p\) ?
solution
The probability is \(\left \vert \langle \phi _{p}|\psi \rangle \right \vert ^{2}\). What goes in the bra is the eigenstate being measured. What goes in the ket is the current state. \begin {align*} \langle \phi _{p}|\psi \rangle & =\int _{-\infty }^{\infty }\langle \phi _{p}|x\rangle \langle x|\psi \rangle dx\\ & =\int _{-\infty }^{\infty }\langle x|\phi _{p}\rangle ^{\ast }\langle x|\psi \rangle dx\\ & =\int _{-\infty }^{\infty }\phi _{p}^{\ast }\relax (x) \psi \left ( x\right ) dx \end {align*}
Now, for the deep well problem for \(0<x<L\), we should know that \(\phi _{p}\relax (x) =\frac {1}{\sqrt {2\pi \hbar }}e^{\frac {ipx}{\hbar }}\) and \(\psi \relax (x) \) will be given. For example \(\psi _{E}\relax (x) =\left \{ \begin {array} [c]{ccc}\sqrt {\frac {2}{L}}\sin \frac {n\pi x}{L} & & 0<x<L\\ 0 & & \text {otherwise}\end {array} \right . \). Hence\[ \langle \phi _{p}|\psi \rangle =\int _{0}^{L}\frac {1}{\sqrt {2\pi \hbar }}e^{\frac {-ipx}{\hbar }}\sqrt {\frac {2}{L}}\sin \frac {n\pi x}{L}dx \] Now evaluate this integral and at the end take the square of the modulus. This will give the probability distribution to measure \(p\). The above was problem 4, in HW7.
Problem says that the system is in some general state \(\psi \relax (x) \) and asks what is the probability distribution to measure position \(x\) ?
solution
The probability is \(\left \vert \langle x|\psi \rangle \right \vert ^{2}\). What goes in the bra is the eigenstate being measured. What goes in the ket is the current state. \begin {align*} \langle x|\psi \rangle & =\int _{-\infty }^{\infty }\langle x|x^{\prime }\rangle \langle x^{\prime }|\psi \rangle dx^{\prime }\\ & =\int _{-\infty }^{\infty }\delta \left (x-x^{\prime }\right ) \psi \left ( x^{\prime }\right ) \rangle dx\\ & =\psi \relax (x) \end {align*}
Hence \(prob\relax (x) =\left \vert \langle x|\psi \rangle \right \vert ^{2}=|\psi \relax (x) |^{2}\)
Problem says that the system is in some general state \(\psi _{E}\left ( x\right ) \) and asks what is the probability distribution to measure position \(x\) ?
solution
The probability is \(\left \vert \langle x|\psi \rangle \right \vert ^{2}\). What goes in the bra is the eigenstate being measured. What goes in the ket is the current or given eigenstate. \begin {align*} \langle x|\psi \rangle & =\int _{0}^{L}\langle x|x^{\prime }\rangle \langle x^{\prime }|\psi \rangle dx^{\prime }\\ & =\int _{0}^{L}\delta \left (x-x^{\prime }\right ) \psi \left (x^{\prime }\right ) dx^{\prime }\\ & =\psi \relax (x) \end {align*}
Hence the probability is \(\left \vert \psi \relax (x) \right \vert ^{2}\). Now, for the deep well problem for \(0<x<L\), we know that \(\psi _{E_{n}}\left ( x\right ) =\left \{ \begin {array} [c]{ccc}\sqrt {\frac {2}{L}}\sin \frac {n\pi x}{L} & & 0<x<L\\ 0 & & \text {otherwise}\end {array} \right . \) then\begin {align*} \left \vert \psi _{E_{n}}\relax (x) \right \vert ^{2} & =\left \vert \sqrt {\frac {2}{L}}\sin \frac {n\pi x}{L}\right \vert ^{2}\\ & =\frac {2}{L}\sin ^{2}\left (\frac {n\pi x}{L}\right ) \end {align*}
Is this correct? Checked, yes correct.
Problem gives that the system is in some general state \(\phi _{p}\left ( x\right ) \) (i.e. momentum eigenstate, not energy eigenstate as above, due to having done momentum measurement done before) and then problem asks what is the probability distribution to measure position \(x\) ?
solution
The probability is \(\left \vert \langle x|\phi _{p}\rangle \right \vert ^{2}\). What goes in the bra is the eigenstate being measured. What goes in the ket is the current eigenstate. \begin {align*} \langle x|\phi _{p}\rangle & =\int _{0}^{L}\langle x|x^{\prime }\rangle \langle x^{\prime }|\phi _{p}\rangle dx^{\prime }\\ & =\int _{0}^{L}\delta \left (x-x^{\prime }\right ) \phi _{p}\left (x^{\prime }\right ) dx^{\prime }\\ & =\phi _{p}\relax (x) \end {align*}
Hence the probability is \(\left \vert \phi _{p}\relax (x) \right \vert ^{2}\). we know that \(\phi _{p}\relax (x) =\frac {1}{\sqrt {2\pi \hbar }}e^{\frac {ipx}{\hbar }}\) then\begin {align*} \left \vert \phi _{p}\relax (x) \right \vert ^{2} & =\left \vert \frac {1}{\sqrt {2\pi \hbar }}e^{\frac {ipx}{\hbar }}\right \vert ^{2}\\ & =\frac {1}{2\pi \hbar } \end {align*}
Which is constant. So if we measure momentum first, then ask for probability of measuring position \(x\) next, it will be the above. Same probability to measure any position? Is this correct? yes.
Problem gives that the system is in some general state \(\phi _{p}\left ( x\right ) \) and asks what is the probability to measure momentum \(p^{\prime }\)?
The probability of measuring momentum \(p^{\prime }\) given that system is already in state \(|\psi _{p}\rangle \equiv \) \(|\phi _{p}\rangle \) is \(\left \vert \langle \phi _{p^{\prime }}|\phi _{p}\rangle \right \vert ^{2}\) where \begin {align*} \langle \phi _{p^{\prime }}|\phi _{p}\rangle & =\int _{-\infty }^{\infty }\langle \phi _{p^{\prime }}|x\rangle \langle x|\phi _{p}\rangle dx\\ & =\int _{-\infty }^{\infty }\langle x|\phi _{p^{\prime }}\rangle ^{\ast }\langle x|\phi _{p}\rangle dx\\ & =\int _{-\infty }^{\infty }\phi _{p^{\prime }}^{\ast }\relax (x) \phi _{p}\relax (x) dx\\ & =\int _{-\infty }^{\infty }\frac {1}{\sqrt {2\pi \hbar }}\exp \left ( \frac {-ip^{\prime }x}{\hbar }\right ) \frac {1}{\sqrt {2\pi \hbar }}\exp \left ( \frac {ipx}{\hbar }\right ) dx\\ & =\frac {1}{2\pi \hbar }\int _{-\infty }^{\infty }\exp \left (\frac {i\left ( p-p^{\prime }\right ) x}{\hbar }\right ) dx \end {align*}
but \(\delta \relax (p) =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{ipx}dx\), therefore \(\delta \left (p-p^{\prime }\right ) =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{i\left (p-p^{\prime }\right ) x}dx\).
Let \(u=\frac {x}{\hbar }\), then \(du=\frac {1}{\hbar }dx\). The integral becomes\begin {align*} \langle \phi _{p^{\prime }}|\phi _{p}\rangle & =\frac {\hbar }{2\pi \hbar }\int _{-\infty }^{\infty }e^{i\left (p-p^{\prime }\right ) u}du\\ & =\frac {1}{2\pi }\left (2\pi \delta \left (p-p^{\prime }\right ) \right ) \\ & =\delta \left (p-p^{\prime }\right ) \end {align*}