# 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: $\sd{X}\sd{P} \ge \frac{\hbar}{2}$