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: $$\psi\rangle\varphi|\psi'\rangle=\langle\varphi|\psi\rangle|\psi\rangle = |\psi''\rangle$$

$$\alpha^\dagger=\alpha^*$$

$$\hat{A}^\dagger = A^{T*}$$

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

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.

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 $$A^{-1}|\psi\rangle=\frac{1}{a}|\psi\rangle$$.

Projection Operator

$$P_n = |\varphi_n\rangle\langle\varphi_n|$$. $$P_n|\psi\rangle=|\varphi_n\rangle\langle\varphi_n|\psi\rangle = c_n|\varphi_n\rangle$$.

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: $$\langle\psi|\psi\rangle = \sum_n c_n^*d_n$$

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 $$A$$)

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

Infinitesimal and Finite Unitary Transformations

Created: 2023-06-25 Sun 02:33

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