If \(\Omega \) is Hermitian operator, then it satisfies\begin {align*} \langle u|\Omega |v\rangle ^{\ast } & =\langle v|\Omega |u\rangle \\ \left (\int u^{\ast }\relax (x) \Omega \left [ v\relax (x) \right ] dx\right ) ^{\ast } & =\int v^{\ast }\relax (x) \Omega \left [ u\relax (x) \right ] dx\\ \int u\relax (x) \Omega \left [ v^{\ast }\relax (x) \right ] dx & =\int v^{\ast }\relax (x) \Omega \left [ u\relax (x) \right ] dx \end {align*}
For this, the boundary terms must vanish. For example, for the operator \(\Omega =-i\frac {d}{dx}\)
\[ \delta \relax (p) =\frac {1}{2\pi }\int _{-\infty }^{\infty }e^{ipx}dx \]
\[ \int _{-\infty }^{\infty }\Psi ^{\ast }\left (x,t\right ) \Psi \left (x,t\right ) dt=1 \]
If a system is in state of \(\Psi \) , then we apply operator \(\hat {A}\), then the average value of the observable quantity is the expectation integral\begin {align*} \left \langle \hat {A}\right \rangle & =\langle \psi |\hat {A}|\psi \rangle \\ & =\frac {\int _{-\infty }^{\infty }\Psi ^{\ast }\hat {A}\Psi dx}{\int _{-\infty }^{\infty }\Psi \Psi dx} \end {align*}
Note that \(\int _{-\infty }^{\infty }\Psi \relax (x) \Psi \relax (x) dx=1\) if the state wave function is already normalized.
Given an operator \(\hat {X}\), acting on \(\Psi \left (x,t\right ) \) then\[ \hat {X}\Psi \left (x,t\right ) =x\Psi \left (x,t\right ) \] The expectation of measuring \(x\) is (assuming everything is normalized)\begin {align*} \left \langle \hat {X}\right \rangle & =\int _{-\infty }^{\infty }\Psi ^{\ast }\left (x,t\right ) \hat {X}\Psi \left (x,t\right ) dx\\ & =\int _{-\infty }^{\infty }\Psi ^{\ast }\left (x,t\right ) x\Psi \left ( x,t\right ) dx\\ & =\left \langle x\right \rangle \end {align*}
Given system is in state \(\psi \relax (x) \). What is the expectation value for \(x\) measurement. Is this same as writing \(\langle X\rangle \). Yes. it is\[ \langle \psi |x|\psi \rangle \]
The probability that position \(x\) of particle is between \(x\) and \(x+dx\) is \(\left \vert \Psi \left (x,t\right ) \right \vert ^{2}dx\). Hence \(\left \vert \Psi \left (x,t\right ) \right \vert ^{2}\) is the probability density.
Note that \begin {align*} \langle \Psi |\Psi \rangle & =\int _{-\infty }^{\infty }\left \vert \Psi \left ( x\right ) \right \vert ^{2}dx\\ \langle \Psi _{1}|\Psi _{2}\rangle & =\int _{-\infty }^{\infty }\Psi _{1}^{\ast }\relax (x) \Psi _{2}\relax (x) dx \end {align*}
Given \(|\Psi \rangle =a|\Psi _{1}\rangle +b|\Psi _{2}\rangle \) then the probabilities to measure \(a\) or \(b\) are\begin {align*} P\relax (a) & =\frac {\left \vert a\right \vert ^{2}}{\left \vert a\right \vert ^{2}+\left \vert b\right \vert ^{2}}\\ P\relax (b) & =\frac {\left \vert b\right \vert ^{2}}{\left \vert a\right \vert ^{2}+\left \vert b\right \vert ^{2}} \end {align*}
| eigenvalue/eigenfunction | \(\hat {x}|x\rangle =x|x\rangle \ \)Where \(x\) is eigenvalue and \(|x\rangle \) is position vector. |
| orthonormal eigenbasis | \(\{|x\rangle \}\rightarrow \left \{ \begin {array} [c]{c}\langle x|x^{\prime }\rangle =\delta \left (x-x^{\prime }\right ) \\ \int _{-\infty }^{\infty }|x\rangle \langle x|\ dx=1 \end {array} \right . \) for \(-\infty <x<\infty \) |
| Vector form to function form | \(\langle x|\psi \rangle \equiv \psi \left ( x\right ) \) probability at position \(x\) |
| Expansion of state vector \(|\psi \rangle \) | \(|\psi \rangle =\int _{-\infty }^{\infty }|x^{\prime }\rangle \langle x^{\prime }|\psi \rangle dx^{\prime }=\int _{-\infty }^{\infty }|x^{\prime }\rangle \psi \left (x^{\prime }\right ) dx^{\prime }\) |
| Eigenfunctions in deep well | Not defined for position operator |
| Operator matrix elements | \(\langle x|\hat {x}|x^{\prime }\rangle =x^{\prime }\delta \left (x-x^{\prime }\right ) \) Operator is diagonal matrix. |
| eigenvalue/eigenfunction | \(\hat {p}|\phi _{p}\rangle =p|\phi _{p}\rangle \) Where \(p\) is eigenvalue and\(\ |\phi _{p}\rangle \) is momentum eigenstate |
| orthonormal eigenbasis | \(\{|\phi _{p}\rangle \}\rightarrow \left \{ \begin {array} [c]{c}\langle \phi _{p}|\phi _{p^{\prime }}\rangle =\delta \left (p-p^{\prime }\right ) \\ \int _{-\infty }^{\infty }|\phi _{p}\rangle \langle \phi _{p}|\ dp=1 \end {array} \right . \) for \(-\infty <p<\infty \) |
| Vector form to function form | \(\langle x|\phi _{p}\rangle \equiv \phi _{p}\left ( x\right ) \) |
| Expansion of state vector \(|\psi \rangle \) | \(|\psi \rangle =\int _{-\infty }^{\infty }|\phi _{p}\rangle \langle \phi _{p}|\psi \rangle dp\) |
| General Eigenfunction | \(\langle x|\phi _{p}\rangle \equiv \phi _{p}\left ( x\right ) =\frac {1}{\sqrt {2\pi \hbar }}\exp \left (\frac {ipx}{\hbar }\right ) \) |
| Operator matrix elements | \(\langle x|\hat {p}|x^{\prime }\rangle =-i\hbar \delta \left (x-x^{\prime }\right ) \frac {d}{dx^{\prime }}\) Operator is not diagonal matrix. |
\[ \hat {H}=\hat {T}+\hat {V}\] Where \(\hat {T}\) is K.E. operator and \(\hat {V}\) is P.E. operator. Recall that \(p=mv\) and \(T=\frac {1}{2}mv^{2}\). Hence \(\hat {T}=\frac {\hat {p}^{2}}{2m}\).
| eigenvalue/eigenfunction | \(\hat {H}|\psi _{E_{n}}\rangle =E_{n}|\psi _{E}\rangle \) Where \(E_{n}\) is eigenvalue (energy level) |
| Orthonormal basis of operator | \(\{|\psi _{E_{n}}\rangle \}\rightarrow \left \{ \begin {array} [c]{c}\langle \psi _{E_{n}}\relax (x) |\psi _{E_{m}}\relax (x) \rangle =\delta \left (E_{n}-E_{m}\right ) \\ \int _{-\infty }^{\infty }|\psi _{E_{n}}\rangle \langle \psi _{E_{n}}|\ dE=1 \end {array} \right . \) for \(n=1,2,\cdots \) (check) |
| Vector form to function form | \(\langle x|\psi _{E_{n}}\rangle \equiv \psi _{E_{n}}\relax (x) \) |
| Expansion of state vector \(|\psi \rangle \) | \(|\psi _{E}\rangle =\sum _{n}|\psi _{E_{n}}\rangle \langle \psi _{E_{n}}|\psi \rangle \) |
| Eigenfunctions for deep well problem | \(\langle x|\psi _{E}\rangle =\psi \left ( x\right ) =\left \{ \begin {array} [c]{ccc}\sqrt {\frac {2}{L}}\sin \left (\frac {n\pi x}{L}\right ) & & 0<x<L\\ 0 & & \text {otherwise}\end {array} \right . \), \(E_{n}=\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}\) |
| Operator matrix elements | \(\langle x|\hat {H}|x^{\prime }\rangle =\frac {1}{2}mv^{2}+V\relax (x) =\delta \left (x-x^{\prime }\right ) \left ( \frac {\hat {p}^{2}}{2m}+\hat {V}\left (x^{\prime }\right ) \right ) =\delta \left (x-x^{\prime }\right ) \left (\frac {-\hbar ^{2}}{2m}\frac {d^{2}}{dx^{\prime 2}}+\hat {V}\left (x^{\prime }\right ) \right ) \) |
The ODE for deep well is derived as follows.\begin {align*} \hat {H}\psi & =E_{n}\psi \\ \left (\hat {T}+\hat {V}\right ) \psi & =E_{n}\psi \end {align*}
But \(\hat {V}=0\) inside and \(\hat {T}=\frac {\hat {p}^{2}}{2m}=\frac {-\hbar ^{2}}{2m}\frac {d^{2}}{dx^{2}}\). Hence the above becomes\begin {align*} \frac {-\hbar ^{2}}{2m}\frac {d^{2}}{dx^{2}}\psi \relax (x) & =E\psi \relax (x) \\ \frac {d^{2}}{dx^{2}}\psi \relax (x) +\frac {2mE}{\hbar ^{2}}\psi \left ( x\right ) & =0\\ \frac {d^{2}}{dx^{2}}\psi \relax (x) +k^{2}\psi \relax (x) & =0 \end {align*}
Where \(k=\sqrt {\frac {2mE}{\hbar ^{2}}}\). The eigenvalues are \(k_{n}\) from solving for boundary conditions at \(x=L\). Now solve as standard second order ODE, with BC \(\psi \relax (0) =0,\psi \relax (L) =0\). The solution becomes\[ \psi \relax (x) =\psi \relax (x) =\left \{ \begin {array} [c]{ccc}\sqrt {\frac {2}{L}}\sin \left (k_{n}x\right ) & & 0<x<L\\ 0 & & \text {otherwise}\end {array} \right . \] Where eigenvalues are \(k_{n}=\frac {n\pi }{L},n=1,2,3,\cdots \)