An operator \(\hat A\) is a rule that maps a ket to another ket of the same space, and a bra to another bra: \[\hat A\left|\phi\right\rangle=\left|\psi\right\rangle,\qquad \left\langle\chi\right|\hat A=\left\langle\phi\right|.\]
Linear operator: \(\hat A\) is linear if
\(\hat A\big(\left|\psi\right\rangle+\left|\phi\right\rangle\big)=\hat A\left|\psi\right\rangle+\hat A\left|\phi\right\rangle\), \(\hat A\big(c\left|\phi\right\rangle\big)=c\hat A\left|\phi\right\rangle\)
\(\big(\hat A+\hat B\big)\left|\psi\right\rangle=\hat A\left|\psi\right\rangle+\hat B\left|\psi\right\rangle\)
\(\hat A\big(a_1\left|\psi_1\right\rangle+a_2\left|\psi_2\right\rangle\big)=a_1\hat A\left|\psi_1\right\rangle+a_2\hat A\left|\psi_2\right\rangle\)
\(\big(\left\langle\psi_1\right|a_1+\left\langle\psi_2\right|a_2\big)\hat A=a_1\left\langle\psi_1\right|\hat A+a_2\left\langle\psi_2\right|\hat A\)
Product: \((\hat A\hat B)\left|\psi\right\rangle=\hat A\big(\hat B\left|\psi\right\rangle\big)\); in general \(\hat A\hat B\ne\hat B\hat A\).
Matrix representation: in a basis \(\{\left|\phi_1\right\rangle,\ldots,\left|\phi_N\right\rangle\}\), \(\hat A\) is an \(N\times N\) matrix with elements \[A_{ij}=\left\langle \phi_i\middle|\hat A\middle|\phi_j\right\rangle .\]
Hermitian Adjoint of an Operator
If \(\hat A\left|\phi\right\rangle=\left|\psi\right\rangle\), then \(\left\langle\phi\right|\hat A^\dagger=\left\langle\psi\right|\); \(\hat A^\dagger\) is the Hermitian adjoint (conjugate) of \(\hat A\).
For a matrix, in two steps:
Step I: transpose, \(\hat A\to\hat A^T\) (rows \(\leftrightarrow\) columns).
Step II: complex-conjugate each element, \(\hat A^\dagger=\big(\hat A^T\big)^*\).
Properties: \[\begin{gathered}
(\hat A^\dagger)^\dagger=\hat A,\qquad (\hat A^n)^\dagger=(\hat A^\dagger)^n,\qquad (a\hat A)^\dagger=a^*\hat A^\dagger,\\
(\hat A+\hat B)^\dagger=\hat A^\dagger+\hat B^\dagger,\qquad (\hat A\hat B)^\dagger=\hat B^\dagger\hat A^\dagger.
\end{gathered}\]
Proof: \((\hat A\hat B)^\dagger=\hat B^\dagger\hat A^\dagger\)
Let \(\hat A\hat B\left|\phi\right\rangle=\left|\psi\right\rangle\Rightarrow\left\langle\phi\right|(\hat A\hat B)^\dagger=\left\langle\psi\right|\).
Write \(\hat B\left|\phi\right\rangle=\left|\chi\right\rangle\Rightarrow\left\langle\phi\right|\hat B^\dagger=\left\langle\chi\right|\).
Then \(\hat A\left|\chi\right\rangle=\left|\psi\right\rangle\Rightarrow\left\langle\chi\right|\hat A^\dagger=\left\langle\psi\right|\Rightarrow\left\langle\phi\right|\hat B^\dagger\hat A^\dagger=\left\langle\psi\right|\).
Comparing: \((\hat A\hat B)^\dagger=\hat B^\dagger\hat A^\dagger\).
Hermitian Operators
\(\hat A\) is Hermitian if \(\hat A^\dagger=\hat A\), i.e. \(\left\langle \phi_i\middle|\hat A\middle|\phi_j\right\rangle =\left\langle \phi_j\middle|\hat A\middle|\phi_i\right\rangle ^*\).
In integral form: \[\int_{-\infty}^{\infty}\psi^*(x)\,\hat A\phi(x)\,dx=\int_{-\infty}^{\infty}\big(\hat A\psi(x)\big)^*\phi(x)\,dx.\]
Examples Based on the Definition of an Operator
Example 2.1 Square of an operator
\(\hat A=\hat D_x+\hat X\), where \(\hat D_x\psi=\dfrac{d\psi}{dx}\) and \(\hat X\psi=x\psi\). Find \(\hat A^2\).
Solution
\(\hat A^2\psi=\Big(\dfrac{d}{dx}+x\Big)\Big(\dfrac{d}{dx}+x\Big)\psi=\dfrac{d^2\psi}{dx^2}+x\dfrac{d\psi}{dx}+\psi+x\dfrac{d\psi}{dx}+x^2\psi\).
\(\hat A^2\psi=\dfrac{d^2\psi}{dx^2}+2x\dfrac{d\psi}{dx}+x^2\psi+\psi\).
\(\Rightarrow\hat A^2=\hat D_x^2+2\hat X\hat D_x+\hat X^2+\hat I\).
Example 2.2 Eigenvalue problem
\(\hat D_x=\dfrac{\partial}{\partial x}\) and \(\psi(x)=A\sin\dfrac{n\pi x}{a}\). (a) Operate \(\hat D_x\) on \(\psi\). (b) Operate \(\hat D_x^2\) on \(\psi\). (c) Which is an eigenvalue problem?
Solution
(a) \(\hat D_x\psi=A\dfrac{n\pi}{a}\cos\dfrac{n\pi x}{a}\) — not proportional to \(\psi\).
(b) \(\hat D_x^2\psi=-\dfrac{n^2\pi^2}{a^2}A\sin\dfrac{n\pi x}{a}=-\dfrac{n^2\pi^2}{a^2}\psi\).
(c) \(\hat D_x^2\) gives an eigenvalue problem, with eigenvalue \(-\dfrac{n^2\pi^2}{a^2}\).
Example 2.3 Momentum eigenfunction
\(\hat P_x=-i\hbar\dfrac{\partial}{\partial x}\). (a) If \(p\) is an eigenvalue, find the eigenfunction. (b) For a particle confined to \(-\frac l2\le x\le\frac l2\), write the normalized eigenstate.
Solution
(a) \(-i\hbar\dfrac{\partial\psi}{\partial x}=p\psi\Rightarrow-i\hbar\displaystyle\int\frac{d\psi}{\psi}=p\int dx\Rightarrow\psi=A\exp\Big(\dfrac{ipx}{\hbar}\Big)\).
(b) \(|A|^2\displaystyle\int_{-l/2}^{l/2}e^{-ipx/\hbar}e^{ipx/\hbar}dx=1\Rightarrow A=\dfrac{1}{\sqrt l}\).
\(\psi(x)=\dfrac{1}{\sqrt l}\exp\Big(\dfrac{ipx}{\hbar}\Big)\) for \(-\frac l2\le x\le\frac l2\); zero otherwise.
Example 2.4 State annihilated by an operator
\(A=i(X^2+1)\dfrac{d}{dx}+iX\) and \(A\psi(x)=0\). Find the normalized \(\psi\).
Solution
\(A\psi=0\Rightarrow\dfrac{d\psi}{dx}=-\dfrac{x}{x^2+1}\psi\Rightarrow\psi(x)=\dfrac{C}{\sqrt{x^2+1}}\).
\(|C|^2\displaystyle\int_{-\infty}^{\infty}\frac{dx}{1+x^2}=|C|^2\pi=1\Rightarrow C=\dfrac{1}{\sqrt\pi}\).
Examples Based on Matrix Representation
Example 2.5 Matrix of an operator
\(\hat S\left|u_1\right\rangle=\left|u_3\right\rangle\), \(\hat S\left|u_2\right\rangle=\left|u_2\right\rangle\), \(\hat S\left|u_3\right\rangle=\left|u_1\right\rangle\), with \(\left\langle u_i\middle|u_j\right\rangle =\delta_{ij}\). Represent \(\hat S\) in the basis \(\{\left|u_1\right\rangle,\left|u_2\right\rangle,\left|u_3\right\rangle\}\).
Solution
Example 2.6 Ladder operators
Lowering operator: \(a\left|n\right\rangle=\sqrt n\left|n-1\right\rangle\); raising operator: \(a^\dagger\left|n\right\rangle=\sqrt{n+1}\left|n+1\right\rangle\); \(n=0,1,2,\ldots\); \(\left\langle m\middle|n\right\rangle =\delta_{mn}\). (a) Represent \(a\), \(a^\dagger\) in the basis \(\left|0\right\rangle,\left|1\right\rangle,\left|2\right\rangle\). (b) Which of \(a\), \(a^\dagger\), \(A=a+a^\dagger\) is Hermitian?
Solution
\(a_{mn}=\left\langle m\middle|a\middle|n\right\rangle =\sqrt n\,\delta_{m,n-1}\) and \(a^\dagger_{mn}=\left\langle m\middle|a^\dagger\middle|n\right\rangle =\sqrt{n+1}\,\delta_{m,n+1}\): \[a=\begin{pmatrix}0&1&0\\0&0&\sqrt2\\0&0&0\end{pmatrix},\qquad
a^\dagger=\begin{pmatrix}0&0&0\\1&0&0\\0&\sqrt2&0\end{pmatrix}.\]
\(a\) and \(a^\dagger\) are not Hermitian, but \(A=a+a^\dagger=\begin{pmatrix}0&1&0\\1&0&\sqrt2\\0&\sqrt2&0\end{pmatrix}\) is Hermitian.
