Joan Lindsay Orr

\( \newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\RR}{\mathbb{R}} \newcommand{\TT}{\mathbb{T}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\H}{\mathcal{H}} \newcommand{\e}{\epsilon} \newcommand{\x}{\mathbf{x}} \newcommand{\y}{\mathbf{y}} \)

\[ \newcommand{\ip}[2]{\langle #1, #2\rangle} \newcommand{\expect}[1]{\langle #1 \rangle} \newcommand{\sd}[1]{\Delta(#1)} \]

Hermitian Operators, Expectations, and Deviations

Let \(H\) be a Hernitian operator on Hilbert space and \(\psi\) be a fixed state vector. Then the expectation of \(H\) with respect to \(\psi\) is defined to be \[ \langle H\rangle_\psi := \langle H\psi, \psi\rangle \]

The motivation of this is to imagine that \(\psi\in L^2(\RR)\) is a normalized wave function. Then \(|\psi(x)|^2\) is a probability density function and if \(h\in L^\infty(\RR)\) is a real-valued function and we write \(H\) for the Hermitian operator of multiplication by \(h\), then \[ \ip{H\psi}{\psi} = \int h(x)\psi(x)\overline{\psi(x)}\, dx = \int h(x) \, |\psi(x)|^2 dx \] is the expectation of \(h\).

In the same vein, we define the standard deviation, \[ \sd{H} := \expect{(H - \expect{H}I)^2}^{1/2} \]

Heisenberg's Inequality

Let \(S\) and \(T\) be Hermitian matrices and \(\psi\) a state vector. Then \[ \Delta(S)\Delta(T) \ge \frac{|\langle [S, T]\psi, \psi\rangle|}{2} \]

Proof. Let \(A\) and \(B\) be two Hermitian operators and \(\psi\) a unit vector. Notice that \[ \langle BA\psi, \psi\rangle = \langle \psi, AB\psi\rangle = \overline{\langle AB\psi, \psi\rangle} \] so that \( \langle [A, B]\psi, \psi\rangle = 2i \mathop{Im}\langle AB\psi, \psi\rangle \). Thus, \[ |\langle [A, B]\psi, \psi\rangle| = 2|\mathop{Im}\langle AB\psi, \psi\rangle| \le 2 |\langle AB\psi, \psi\rangle| = 2 |\langle B\psi, A\psi\rangle| \] By the Cauchy-Schwartz Inequality, \( |\langle B\psi, A\psi\rangle| \le \|A\psi\| \|B\psi\| \) and so \[ |\langle [A, B]\psi, \psi\rangle| \le 2 \|A\psi\| \|B\psi\| = 2 \langle A^2 \psi, \psi\rangle^\frac{1}{2} \langle B^2 \psi, \psi\rangle^\frac{1}{2} \] Now if \(S\) and \(T\) are Hermitian operators, take \(A = S - \langle S\psi, \psi \rangle I\) and \(B = T - \langle T\psi, \psi \rangle I\). Note that \([S, T] = [A, B]\) and so \[ |\langle [S, T]\psi, \psi\rangle \le 2 \langle A^2\psi, \psi\rangle^\frac{1}{2}\langle B^2\psi, \psi\rangle^\frac{1}{2} \] Now \(\langle S\psi, \psi \rangle\) is the expected outcome of measuring qubits in state \(\psi\) and so \(\langle A^2\psi, \psi \rangle\) is the variance. Writing \(\Delta(S)\) (resp. \(\Delta(T)\)) for the standard deviation, we obtain \[ \Delta(S)\Delta(T) \ge \frac{|\langle [S, T]\psi, \psi\rangle|}{2} \] and the result follows.

Let \(\psi\) be a smooth \(L^2\) function. The poisition operator is \(X := M_x\) (multiplication by \(x\)) and the momentum operator is \[ P := -i\hbar\frac{\partial}{\partial x} \] Clearly \[ XP\psi(x) = -i\hbar x\frac{\partial\psi}{\partial x}(x) \] and \[ PX\psi(x) = -i\hbar \frac{\partial}{\partial x}x\psi(x) = -i\hbar \psi(x) -i\hbar x\frac{\partial\psi}{\partial x}(x) \] Thus \[ [X, P]\psi = i\hbar\psi \qquad\text{and}\qquad [X, P] = i\hbar I \] From this we get Heisenberg's Inequality for position and momentum: \[ \sd{X}\sd{P} \ge \frac{\hbar}{2} \]