5.17 Quantum mechanics cheat sheet

5.17.1 Hermitian operator in function spaces

If $\mathrm{\Omega }$ is Hermitian operator, then it satisfies$$\begin{array}{rl}⟨u|\mathrm{\Omega }|v{⟩}^{\ast }& =⟨v|\mathrm{\Omega }|u⟩\\ {(\int u^{\ast }(x)\mathrm{\Omega }[v(x)]dx)}^{\ast }& =\int v^{\ast }(x)\mathrm{\Omega }[u(x)]dx\\ \int u(x)\mathrm{\Omega }[v^{\ast }(x)]dx& =\int v^{\ast }(x)\mathrm{\Omega }[u(x)]dx\end{array}$$

For this, the boundary terms must vanish. For example, for the operator $\mathrm{\Omega }=-i\frac{d}{dx}$

5.17.2 Dirac delta relation to integral

$$\delta (p)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{ipx}dx$$

5.17.3 Normalization condition

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{\Psi }}^{\ast }(x,t)\mathrm{\Psi }(x,t)dt=1$$

5.17.4 Expectation (or average value)

If a system is in state of $\mathrm{\Psi }$ , then we apply operator $\hat{A}$, then the average value of the observable quantity is the expectation integral$$\begin{array}{rl}⟨\hat{A}⟩& =⟨\psi |\hat{A}|\psi ⟩\\ & =\frac{{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{\Psi }}^{\ast }\hat{A}\mathrm{\Psi }dx}{{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Psi }\mathrm{\Psi }dx}\end{array}$$

Note that ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Psi }(x)\mathrm{\Psi }(x)dx=1$ if the state wave function is already normalized.

Given an operator $\hat{X}$, acting on $\mathrm{\Psi }(x,t)$ then$$\hat{X}\mathrm{\Psi }(x,t)=x\mathrm{\Psi }(x,t)$$ The expectation of measuring $x$ is (assuming everything is normalized)$$\begin{array}{rl}⟨\hat{X}⟩& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{\Psi }}^{\ast }(x,t)\hat{X}\mathrm{\Psi }(x,t)dx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{\Psi }}^{\ast }(x,t)x\mathrm{\Psi }(x,t)dx\\ & =⟨x⟩\end{array}$$

Given system is in state $\psi (x)$. What is the expectation value for $x$ measurement. Is this same as writing $⟨X⟩$. Yes. it is$$⟨\psi |x|\psi ⟩$$

5.17.5 Probability

The probability that position $x$ of particle is between $x$ and $x+dx$ is ${\left|\mathrm{\Psi }(x,t)\right|}^{2}dx$. Hence ${\left|\mathrm{\Psi }(x,t)\right|}^2$ is the probability density.

Note that $$\begin{array}{rl}⟨\mathrm{\Psi }|\mathrm{\Psi }⟩& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|\mathrm{\Psi }\left(x\right)\right|}^{2}dx\\ ⟨{\mathrm{\Psi }}_1|{\mathrm{\Psi }}_2⟩& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{\Psi }}_1^{\ast }(x){\mathrm{\Psi }}_2(x)dx\end{array}$$

Given $|\mathrm{\Psi }⟩=a|{\mathrm{\Psi }}_1⟩+b|{\mathrm{\Psi }}_2⟩$ then the probabilities to measure $a$ or $b$ are$$\begin{array}{rl}P(a)& =\frac{{\left|a\right|}^2}{{\left|a\right|}^2+{\left|b\right|}^2}\\ P(b)& =\frac{{\left|b\right|}^2}{{\left|a\right|}^2+{\left|b\right|}^2}\end{array}$$

5.17.6 Position operator $\hat{x}$

eigenvalue/eigenfunction	$\hat{x}|x⟩=x|x⟩$Where $x$ is eigenvalue  and $|x⟩$ is position vector.
orthonormal eigenbasis	$\{|x⟩\}\to \{\begin{array}{c}⟨x|x^{\prime }⟩=\delta (x-x^{\prime })\\ {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|x⟩⟨x|dx=1\end{array}$ for $-\mathrm{\infty }<x<\mathrm{\infty }$
Vector form to function form	$⟨x|\psi ⟩\equiv \psi \left(x\right)$ probability at position $x$
Expansion of state vector $|\psi ⟩$	$|\psi ⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|x^{\prime }⟩⟨x^{\prime }|\psi ⟩d{x}^{\prime }={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|x^{\prime }⟩\psi \left(x^{\prime }\right)d{x}^{\prime }$
Eigenfunctions in deep well	Not defined for position operator
Operator matrix elements	$⟨x|\hat{x}|x^{\prime }⟩=x^{\prime }\delta (x-x^{\prime })$ Operator is diagonal matrix.

5.17.7 Momentum operator $\hat{p}$

eigenvalue/eigenfunction	$\hat{p}|{\varphi }_p⟩=p|{\varphi }_p⟩$ Where $p$ is eigenvalue  and$|{\varphi }_p⟩$ is momentum eigenstate
orthonormal eigenbasis	$\{|{\varphi }_p⟩\}\to \{\begin{array}{c}⟨{\varphi }_p|{\varphi }_{p^{\prime }}⟩=\delta (p-p^{\prime })\\ {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|{\varphi }_p⟩⟨{\varphi }_p|dp=1\end{array}$ for $-\mathrm{\infty }<p<\mathrm{\infty }$
Vector form to function form	$⟨x|{\varphi }_p⟩\equiv {\varphi }_p\left(x\right)$
Expansion of state vector $|\psi ⟩$	$|\psi ⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|{\varphi }_p⟩⟨{\varphi }_p|\psi ⟩dp$
General Eigenfunction	$⟨x|{\varphi }_p⟩\equiv {\varphi }_p\left(x\right)=\frac{1}{\sqrt{2\pi \hslash }}\exp \left(\frac{ipx}{\hslash }\right)$
Operator matrix elements	$⟨x|\hat{p}|x^{\prime }⟩=-i\hslash \delta (x-x^{\prime })\frac{d}{d{x}^{\prime }}$ Operator is not diagonal matrix.

5.17.8 Hamilitonian operator $\hat{H}$

$$\hat{H}=\hat{T}+\hat{V}$$ Where $\hat{T}$ is K.E. operator and $\hat{V}$ is P.E. operator. Recall that $p=mv$ and $T=\frac{1}{2}m{v}^2$. Hence $\hat{T}=\frac{{\hat{p}}^2}{2m}$.  

eigenvalue/eigenfunction	$\hat{H}|{\psi }_{E_n}⟩=E_n|{\psi }_E⟩$ Where $E_n$ is eigenvalue (energy level)
Orthonormal basis of operator	$\{|{\psi }_{E_n}⟩\}\to \{\begin{array}{c}⟨{\psi }_{E_n}(x)|{\psi }_{E_m}(x)⟩=\delta (E_n-E_m)\\ {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|{\psi }_{E_n}⟩⟨{\psi }_{E_n}|dE=1\end{array}$ for $n=1,2,\cdots$ (check)
Vector form to function form	$⟨x|{\psi }_{E_n}⟩\equiv {\psi }_{E_n}(x)$
Expansion of state vector $|\psi ⟩$	$|{\psi }_E⟩=\sum _n|{\psi }_{E_n}⟩⟨{\psi }_{E_n}|\psi ⟩$
Eigenfunctions for deep well problem	$⟨x|{\psi }_E⟩=\psi \left(x\right)=\{\begin{array}{ccc}\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)& & 0<x<L\\ 0& & \text{otherwise}\end{array}$, $E_n=\frac{n^2{\pi }^2{\hslash }^2}{2m{L}^2}$
Operator matrix elements	$⟨x|\hat{H}|x^{\prime }⟩=\frac{1}{2}m{v}^2+V(x)=\delta (x-x^{\prime })(\frac{{\hat{p}}^2}{2m}+\hat{V}\left(x^{\prime }\right))=\delta (x-x^{\prime })(\frac{-{\hslash }^2}{2m}\frac{d^2}{d{x}^{\prime 2}}+\hat{V}\left(x^{\prime }\right))$

The ODE for deep well is derived as follows.$$\begin{array}{rl}\hat{H}\psi & =E_n\psi \\ (\hat{T}+\hat{V})\psi & =E_n\psi \end{array}$$

But $\hat{V}=0$ inside and $\hat{T}=\frac{{\hat{p}}^2}{2m}=\frac{-{\hslash }^2}{2m}\frac{d^2}{d{x}^2}$. Hence the above becomes$$\begin{array}{rl}\frac{-{\hslash }^2}{2m}\frac{d^2}{d{x}^2}\psi (x)& =E\psi (x)\\ \frac{d^2}{d{x}^2}\psi (x)+\frac{2mE}{{\hslash }^2}\psi \left(x\right)& =0\\ \frac{d^2}{d{x}^2}\psi (x)+k^2\psi (x)& =0\end{array}$$

Where $k=\sqrt{\frac{2mE}{{\hslash }^2}}$. The eigenvalues are $k_n$ from solving for boundary conditions at $x=L$.  Now solve as standard second order ODE, with BC $\psi (0)=0,\psi (L)=0$. The solution becomes$$\psi (x)=\psi (x)=\{\begin{array}{ccc}\sqrt{\frac{2}{L}}\sin \left(k_{n}x\right)& & 0<x<L\\ 0& & \text{otherwise}\end{array}$$ Where eigenvalues are $k_n=\frac{n\pi }{L},n=1,2,3,\cdots$