Quantum Mechancial Operators

Definition

An operator \(\hat{A}\), acting on a state, is the mathematical rule that \(\hat{A}|\psi\rangle=|\psi'\rangle\) and \(\langle\varphi|A=\langle\varphi'|\).

Linear Operators

We will use linear operators for PH651.

Forbidden Quantities

\(A\langle\varphi|\)
\(|\varphi\rangle A\)

Properties

Expectation values (mean)

\(\overline{A}=\langle\psi|\hat{A}|\psi=\langle\hat{A}\rangle\) for normalized \(|\psi\rangle\).

For non-normalized, \(\langle\hat{A}\rangle=\frac{\langle\psi|A|\psi\rangle}{\langle\psi|\psi\rangle}\).

Outer Product Operator

\(|\varphi\rangle\langle\psi|\)

Applying: \(\psi\rangle\varphi|\psi'\rangle=\langle\varphi|\psi\rangle|\psi\rangle = |\psi''\rangle\)

Hermitian Adjoint (conjugate)

\(\alpha^\dagger=\alpha^*\)

\(\hat{A}^\dagger = A^{T*}\)

\(\langle\psi|\hat{A}^\dagger|\varphi\rangle = \langle\varphi|A|\psi\rangle^*\)

Hermitian Operator

\(A^\dagger = A\). \(\langle\psi|\hat{A}|\varphi\rangle = \rangle\varphi|A|\psi\rangle^*\).

Antihermitian Operator

\(B^\dagger = -B\). \(\langle\psi|\hat{A}|\varphi\rangle = -\rangle\varphi|A|\psi\rangle^*\).

Examples

\(\hat{A}^{\dagger\dagger}\)
\((\alpha\hat{A})^\dagger = \alpha^*A^\dagger\)
\((A+B+C+D)^\dagger = (A^\dagger + B^\dagger + C^\dagger + D^\dagger)\)
\((ABCD)^\dagger=D^\dagger C^\dagger B^\dagger A^\dagger\)
\((ABCD|\varphi\rangle)^\dagger = \langle\varphi|D^\dagger C^\dagger B^\dagger A^\dagger\)

Hermicity

\(\hat{A}=\hat{X}\) is it Hermitian? \(\langle\psi|\hat{X}|\varphi\rangle=\int \psi^*(x)x\varphi(x)dx = (\int\psi(x)x^*\varphi^*(x)dx)^* = (\int\varphi^*(x)x\psi(x)dx)^* = \langle\varphi|\hat{X}|\varphi\rangle^*\), Yes.
Is \(\hat{A}=\frac{d}{d\hat{X}}\) Hermitian? \(\langle\psi|\frac{d}{d \hat{X}}|\varphi\rangle = \int\psi^*(x) \frac{d}{dx}\varphi(x)dx = 0 - \int\varphi^*(x)\frac{d\psi}{dx}dx^* = -\langle\varphi|\hat{A}|\psi\rangle^*\), No, anti-hermititan.
\(-i\hbar\frac{d}{d x}\) is it Hermitian? Yes.

Commutator

Functions of Operators

Inverse

If it exists, then \(A^{-1}A=AA^{-1}=\mathbb{I}\).

If \(A|\psi\rangle = a|\psi\rangle\) then \(A^{-1}a|\psi\rangle = |\psi\rangle\), or \(A^{-1}|\psi\rangle=\frac{1}{a}|\psi\rangle\).

Unitary Operators

Projection Operator

Thus, \(\mathbb{I} = \sum_i P_i\).

Matrix Representations of Kets, Bras, Operators

Operators: Matrix - Product of Column Vector with Row Vector \(A^i_j = \langle i|A|j\rangle, (A^i_j)^\dagger = \langle j|A^\dagger|i\rangle\)
Kets: Column Vector. \(|\psi\rangle = \sum_n c_n |\varphi_n\rangle=\begin{pmatrix}\langle\varphi_1|\psi\rangle\\\cdots\end{pmatrix}=\begin{pmatrix}c_1\\c_2\\\cdots\end{pmatrix}\)
Bras: Row Vector. \(\langle\psi| = (\langle\psi|\varphi_1\rangle \cdots) = (c_1^* \cdots)\)
Inner Product: \(\langle\psi|\psi\rangle = \sum_n c_n^*d_n\)

Trace

Types

Real Matrix

\(A=A^*\)

Imaginary Matrix

\(A=-A^*\)

Symmetric Matrix

\(A^T=A\)

Anti-Symmetric Matrix

\(A^T=-A\)

N.B. The diagonal of an Anti-Symmetric Matrix must be zero.

Orthogonal Matrix

\(A^T=A^{-1}\)

Theorem

A Real-Orthogonal Matrix is Unitary.

Expectation Values

\(\langle A\rangle = \sum_{n,m}\langle\psi|\varphi_n\rangle\langle\varphi_n|A|\varphi_m\rangle\langle\varphi_m|\psi\rangle = \sum_{n,m} a_n^*a_mA^n_m\). Weighted average.

For Diagonal Operator, \(\langle A\rangle = \sum_n A_n |a_n|^2\). (In this case, we expect diagonalization since \(|\varphi_n\rangle\) are orthonormal bases for \(A\))

Quantum Mechanical Measurements

Quantum Mechanical Observables

Uncertainty Principle

\((\Delta A)^2(\Delta B)^2=\langle (\Delta \hat{A})^2\rangle\langle(\Delta \hat{B})^2\rangle \geq \frac{1}{4}|\langle [A,B]\rangle|^2\)

\(\Delta \hat{A} = \hat{A} - \langle \hat{A} \rangle\).
\(\Delta A = \sqrt{\langle \hat{A}^2\rangle - \langle \hat{A}\rangle^2}\)
\((\Delta \hat{A})^2 = (\hat{A}-\langle \hat{A}\rangle)^2\).
\(\langle (\Delta A)^2 \rangle = \langle \hat{A}^2\rangle - \langle \hat{A}\rangle^2\).
\(|a\rangle = \Delta \hat{A}|\psi\rangle\)
\(|b\rangle = \Delta \hat{B}|\psi\rangle\)
\(|\langle a|b\rangle|^2\leq\langle a|a\rangle\langle b|b\rangle = \langle \psi|\Delta \hat{A}^\dagger \Delta \hat{A}|\psi\rangle \langle \psi|\Delta \hat{B}^\dagger \Delta \hat{B}|\psi\rangle = \langle \hat{A}^2\rangle \langle \hat{B}^2\rangle = (\Delta A)^2(\Delta B)^2\)
\(|\langle a|b\rangle = \langle (\Delta \hat{A}\Delta\hat{B})\rangle\)
\(|\langle (\Delta\hat{A})(\Delta\hat{B})\rangle|^2\leq(\Delta A)^2(\Delta B)^2\)
\(AB = \frac{1}{2}([A,B] + \{A,B\}) \Rightarrow \Delta \hat{A}\Delta\hat{B} = \frac{1}{2}([\Delta \hat{A},\Delta\hat{B}]+\{\Delta \hat{A},\Delta\hat{B}\}) = \frac{1}{2}([\hat{A},\hat{B}] + \{\Delta\hat{A},\Delta\hat{B}\})\)
Anti-commutator has real expectation value and expectation value of commutator is imaginary.
\(\langle \Delta\hat{A}\Delta\hat{B}\rangle^2 = \frac{1}{4}(|\langle [\hat{A},\hat{B}]\rangle|^2 + |\langle \{\Delta\hat{A},\Delta\hat{B}\}|^2)\geq \frac{1}{4}|\langle[\hat{A},\hat{B}]\rangle|^2\)
\((\Delta A)(\Delta B)\geq \frac{1}{2}|\langle [\hat{A},\hat{B}]\rangle|\)