06. Statistical Aspects of Quantum Theory

6.1 Expectation and Dispersion

Expectation Value of Operator $A$

Consider normalised quantum state $\psi$ then

Expectation Value of an Operator $A$ is

$${\mathbb{E}}_{\psi }(A)\equiv ⟨\psi \mid A\psi ⟩$$

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Expectation Value of Function of Position Operator $X$

Consider wave function $\psi =\Psi (x,t)$ in one dimension
Let $A=f(X)$ be a function of position operator $X$ then

$${\mathbb{E}}_{\psi }(f(X))={\int }_{-\infty }^{\infty }\overline{\Psi (x,t)}f(x)\Psi (x,t)dx={\int }_{-\infty }^{\infty }f(x)|\Psi (x,t){|}^{2}dx$$

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Expectation Value of Operator $A$ with Complete Orthonormal Basis of Eigenstates

Suppose $A$ has complete orthonormal basis of eigenstates ${\psi }_n$ where

$$A{\psi }_n={\alpha }_n{\psi }_n\text{ and }\psi =\sum _n{c}_n{\psi }_n$$

Then

$${\mathbb{E}}_{\psi }(A)=⟨\psi \mid A\psi ⟩=\sum _n{c}_n⟨\psi \mid A{\psi }_n⟩=\sum _n{\alpha }_n{c}_n⟨\psi \mid {\psi }_n⟩=\sum _n{\alpha }_n|c_n{|}^2$$

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Identity Operator

Identity Operator $1$ defined by

$$1\psi =\psi \text{ for all }\psi \in \mathcal{H}$$

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Non-Negative Operator

Operator $A$ is non-negative if

$$⟨\psi \mid A\psi ⟩\ge 0\text{ for all }\psi \in \mathcal{H}$$

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Properties of Expectation Operator

Let $A,B$ be operators then

1. Linearity

$${\mathbb{E}}_{\psi }(\alpha A+\beta B)=\alpha {\mathbb{E}}_{\psi }(A)+\beta {\mathbb{E}}_{\psi }(B)\text{ for all }\alpha ,\beta \in \mathbb{C}$$

2. Expectation of Identity is $1$

$${\mathbb{E}}_{\psi }(1)=1$$

3. Expectation of Self-Adjoint

$${\mathbb{E}}_{\psi }(A)\in \mathbb{R}$$

4. Let $A$ be a non-negative operator then

$${\mathbb{E}}_{\psi }(A)\ge 0$$

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Dispersion of Observable

Let $A$ be an observable then

Dispersion of Observable $A$ with normalise quantum state $\psi$ defined as

$${\Delta }_{\psi }(A)\equiv \sqrt{{\mathbb{E}}_{\psi }((A-{\mathbb{E}}_{\psi }(A)1{)}^2)}\equiv \sqrt{{\mathbb{E}}_{\psi }(A^2)-({\mathbb{E}}_{\psi }(A){)}^2}$$

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Dispersion of Observable $A$ with Eigenstate

Consider dispersion of observable $A$ with a normalised state $\varphi$ then

$${\Delta }_{\psi }(A)=0⟺\psi \text{ is an eigenstate of }A$$

where $\varphi$ has associated eigenvalue ${\mathbb{E}}_{\psi }(A)$

Proof

Suppose $\psi$ is eigenstate of $A$ with eigenvalue $\alpha$ so $A\psi =\alpha \psi$ then

$${\mathbb{E}}_{\psi }(A)=⟨\psi \mid A\psi ⟩=⟨\psi \mid \alpha \psi ⟩=\alpha ⟨\psi \mid \psi ⟩=\alpha$$

And

$$\begin{array}{rl}||(A-{\mathbb{E}}_{\psi }(A)1)\psi |{|}^2& =⟨(A-{\mathbb{E}}_{\psi }(A)1)\psi \mid (A-{\mathbb{E}}_{\psi }(A)1)\psi ⟩\\ & =⟨\psi \mid (A-{\mathbb{E}}_{\psi }(A)1{)}^2\psi ⟩\text{(by self-adjointness)}\\ & ={\Delta }_{\psi }(A{)}^2\end{array}$$

Hence ${\Delta }_{\psi }(A)=0$ if and only $A\psi =E_{\psi }(A)\psi$ if and only if $\psi$ is an eigenstate of $A$ with eigenvalue ${\mathbb{E}}_{\psi }(A)$

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6.2 Commutation Relation

Commutator of Two Operators

Let $A$ and $B$ be two operators

Commutator is defined as

$$[A,B]\equiv AB-BA$$

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Canonical Commutation Relations

In one dimensions then

$$[X,P]=i\hslash 1$$

In three dimensions then

$$[X_i,P_j]=i\hslash {\delta }_{ij}1,[X_i,X_j]=0=[P_i,P_j]$$

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Properties of Commutator

Let $A,B,C$ be operators then commutators satisfy

1. Anti-Symmetry

$$[A,B]=-[B,A]$$

2. Linearity

$$[A,\beta B+\gamma C]=\beta [A,B]+\gamma [A,C]\text{ for all }\beta ,\gamma \in \mathbb{C}$$

3. Leibniz Rule

$$[A,BC]=B[A,C]+[A,B]C\text{ and }[AB,C]=A[B,C]+[A,C]B$$

4. Jacobi Identity

$$[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$$

5. If $[A,B]=iC$ with $A,B$ both self-adjoint then

$$C\text{ is also self-adjoint}$$

Proof

Proof $(iii)$

$$[A,BC]=ABC-BCA=(AB-BA)C+B(AC-CA)=B[A,C]+[A,B]C$$

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6.3 Heisenberg’s Uncertainty Principle

Extra Properties of Self-Adjoint Operators

Let $A,B$ be self-adjoint operators and suppose $[A,B]=iC$

For $s\in \mathbb{R}$ and normalised $\varphi \in \mathcal{H}$ then

1. $(A-isB{)}^{\ast }(A-isB)=A^2+sC+s^2{B}^2$

2. $||(A-isB)\psi |{|}^2={\mathbb{E}}_{\psi }(A^2)+s{\mathbb{E}}_{\psi }(C)+s^2{\mathbb{E}}_{\psi }(B^2)$

3. ${\mathbb{E}}_{\psi }(A^2){\mathbb{E}}_{\psi }(B^2)\ge \frac{1}{4}({\mathbb{E}}_{\psi }(C){)}^2$

Note that $(3)$ holds with equality if and only if exists $s_0\in \mathbb{R}$ such that $(A-i{s}_{0}B)\psi =0$

Proof

$(1)$

$$\begin{array}{rl}(A-isB{)}^{\ast }(A-isB)& =(A+isB)(A-isB)\\ & =A^2-is(AB-BA)+s^2{B}^2\\ & =A^2+sC+s^2{B}^2\end{array}$$

$(2)$

$$\begin{array}{rl}||(A-isB)\psi |{|}^2& =⟨(A-isB)\psi \mid (A-isB)\psi ⟩\\ & =⟨\psi \mid (A-isB{)}^{\ast }(A-isB)\psi ⟩\\ & =⟨\psi \mid (A^2+sC+s^2{B}^2)\psi ⟩\\ & ={\mathbb{E}}_{\psi }(A^2)+s{\mathbb{E}}_{\psi }(C)+s^2{\mathbb{E}}_{\psi }(B^2)\end{array}$$

$(3)$
For $(2)$ the right hand side is a quadratic and left side is $\ge 0$ so discriminant is $\le 0$ so

$$({\mathbb{E}}_{\psi }(C){)}^2-4{\mathbb{E}}_{\psi }(A^2){\mathbb{E}}_{\psi }(B^2)\le 0$$

Quadratic has repeated real root as $s=s_0$ so

$$0={\mathbb{E}}_{\psi }(A^2)+s{\mathbb{E}}_{\psi }(C)+s^2{\mathbb{E}}_{\psi }(B^2)=||(A-isB)\psi |{|}^2$$

Hence $s_0$ is a root if and only if $(A-i{s}_{0}B)\psi =0$

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Heisenberg's Uncertainty Principle corollary

Suppose $\psi$ is normalised then

$${\Delta }_{\psi }(X){\Delta }_{\psi }(P)\ge \frac{1}{2}\hslash$$

with equality if and only if

$$\psi (x)=\exp \left[\frac{1}{2s_0\hslash }(x-\mu {)}^2+\gamma \right]$$

where negative constant $s_0$ and complex constants $\mu ,\gamma \in \mathbb{C}$

Proof

Define

$$A\equiv X-{\mathbb{E}}_{\psi }(X)1\text{ and }B\equiv P-{\mathbb{E}}_{\psi }(P)1$$

so by linearity of commutator then

$$[A,B]=[X,P]-{\mathbb{E}}_{\psi }(P)[X,1]-{\mathbb{E}}_{\psi }(X)([1,P-{\mathbb{E}}_{\psi }(P)1])=[X,P]=i\hslash 1$$

Hence

$$C=-i[A,B]-\hslash 1$$

Hence

$${\mathbb{E}}_{\psi }(A^2)=({\Delta }_{\psi }(X){)}^2\text{ and }{\mathbb{E}}_{\psi }(B^2)=({\Delta }_{\psi }(P){)}^2$$

With ${\mathbb{E}}_{\psi }(C)=\hslash$ by Extra properties of self-adjoint operators

For equality from $(iii)$ Extra properties of self-adjoint operators then

Exists real $s_0$ with $(A-i{s}_{0}B)\psi =0$ so
Let $\mu ={\mathbb{E}}_{\psi }(X)-i{s}_0{\mathbb{E}}_{\psi }(P)$ and substitution with $P=-\hslash \frac{d}{dx}$ then

$$0=(A-i{s}_{0}B)\psi =\left(x-i{s}_0(-\hslash )\frac{d}{dx}-\mu \right)\psi$$

Rearranging to

$${\psi }^{\prime }=\frac{2}{s_0\hslash }(x-\mu )\psi$$

Hence

$$\psi (x)=\exp \left(\frac{1}{2s_0\hslash }(x-\mu {)}^2+\gamma \right)$$

Require $s_0<0$ for gaussian integral

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Generalised Uncertainty Principle corollary

Let $A_1,A_2$ be self-adjoint operators then
Let $\psi \in \mathcal{H}$ be a normalised

$${\Delta }_{\psi }(A_1){\Delta }_{\psi }(A_2)\ge \frac{1}{2}|{\mathbb{E}}_{\psi }(-i[A_1,A_2])|$$

Proof

Similar to proof for Heisenberg’s uncertainty principle where

$$A\equiv A_1-{\mathbb{E}}_{\psi }(A_1)1\text{ and }B\equiv A_2-{\mathbb{E}}_{\psi }(A_2)1$$

where $[A,B]=[A_1,A_2]$

By definition of dispersion then

$${\mathbb{E}}_{\psi }(A^2)=({\Delta }_{\psi }(A_1){)}^2\text{ and }{\mathbb{E}}_{\psi }(B^2)=({\Delta }_{\psi }(A_2){)}^2$$

Taking square root of $(iii)$ inequality for Extra properties of self-adjoint operators gives result

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