Expectation values in quantum mechanics

We had seen that the square of the wave function, $\vert\Psi({x},t)\vert^2$, describing a particle in a potential $V({x})$, is a probability density to find the particle at ${x}$ at time $t$. The result of a single measurement of ${x}$ can only be predicted to have a certain probability, but if many measurements of the position ${x}$ under identical conditions are repeated, the average value (expectation value) of ${x}$ is

$$\langle x \rangle_t =\int dx \vert\Psi({x},t)\vert^2 x.$$ (58)

We have again adopted a one-dimensional version for simplicity, in three dimensions the expectation value is completely analogous,

$$\langle {\bf x} \rangle_t =\int d^3x \vert\Psi({\bf x},t)\vert^2 {\bf x}.$$ (59)

Note that the expectation value is now a three-dimensional vector which is a function of time $t$. We have indicated the time-dependence by the notation $\langle...\rangle_t$.

Next, we would like to know the expectation value of the momentum ${\bf p}$ of the particle. To determine ${\bf x}$ and ${\bf p}$ for a given massive object (like a planet revolving around the sun) at time $t$ is one of the aims of classical mechanics. In quantum mechanics, we only have the probability density $\vert\Psi({x},t)\vert^2$, but we can calculate expectation values: We define the expectation value for the momentum $p$ (one-dimensional version) as

$$\langle p \rangle_t \equiv m\frac{d}{dt}\langle x \rangle_t .$$ (60)

This seems plausible because in classical mechanics $p=m\dot{x}$. Later we will see that an equivalent definition is also possible, using the de Broglie relation $p=\hbar k$. We write

$$\langle p \rangle_t = m\frac{d}{dt}\langle x \rangle_t = m \int dx \frac{d}{dt}\vert\Psi({x},t)\vert^2 x= -m\int dx \frac{\partial}{\partial x} j(x,t) x .$$ (61)

Now, we re-call the definition of the current-density and the continuity equation, Eq.(1.39),

$$\fbox{$ \begin{array}{rcl} \displaystyle & &\frac{\partial}{\part... ...si(x,t) - \Psi(x,t)\frac{\partial}{\partial x}\Psi^*(x,t)\right]. \end{array}$}$$

We therefore find

$$\langle p \rangle_t = {\mbox{\rm [partial integration]}} = m \int dx j(x,t) ={\mbox{\rm [definition of j]}}$$

$$= -\frac{i\hbar}{2} \int dx [\Psi^*({x},t)\partial_x \Psi({x},t)-\Psi({x},t)\partial_x \Psi^*({x},t)] ={\mbox{\rm [partial integration]}}$$

$$= -\frac{i\hbar}{2} \int dx [\Psi^*({x},t)\partial_x \Psi({x},t)+\partial_x\Psi({x},t) \Psi^*({x},t)]$$

$$= \int dx [\Psi^*({x},t)\frac{\hbar \partial_x}{i} \Psi({x},t)].$$ (62)

We compare

$$\langle x \rangle_t =\int dx \Psi^*({x},t) x \Psi({x},t),\quad \langle p \rangle_t =\int dx \Psi^*({x},t) \frac{\hbar \partial_x}{i} \Psi({x},t)$$ (63)

and recognize that the position $x$ corresponds to the (somewhat trivial) operator `multiplication with $x$'. On the other hand, the momentum corresponds to the completely non-trivial operator $-i\hbar \partial_x$. A similar calculation leads to

$$\langle x^2 \rangle_t =\int dx \Psi^*({x},t) x^2 \Psi({x},t),\qua... ...t =\int dx \Psi^*({x},t) \left[\frac{\hbar \partial_x}{i}\right]^2 \Psi({x},t).$$ (64)

Again, the above can easily be generalized to three dimensions when $x\to {\bf x}$ and $\partial_x\to (\partial_x,\partial_y,\partial_z)=\nabla$ (gradient or Nabla-operator).

Axiom 2: Expectation values of functions $F({\bf x})$ of the position or $G({\bf p})$ of the momentum for a quantum mechanical system described by a wave function $\Psi({\bf x},t)$ are calculated as

$$\langle F({\bf x}) \rangle_t = \int d^3x \Psi^*({\bf x},t) F({\bf x}) \Psi({\bf x},t)$$

$$\langle G({\bf p}) \rangle_t = \int d^3x \Psi^*({\bf x},t) G\left(\frac{\hbar \nabla}{i}\right) \Psi({\bf x},t).$$ (65)

The position ${\bf x}$ corresponds to the operator `multiplication with ${\bf x}$', the momentum ${\bf p}$ to the operator $-i\hbar \nabla$, applied to the wave function as in (1.63),(1.64),(1.65).

This correspondence in particular holds for the total energy, which in classical mechanics for a conservative system (energy is conserved) is given by a Hamilton function

$$H({\bf p},{\bf x})=\frac{{\bf p}^2}{2m}+V({\bf x}).$$ (66)

The correspondence principle from axiom 2 tells us that this Hamilton function in quantum mechanics has to be replaced by a Hamilton operator (Hamiltonian) $\hat{H}$

$$\hat{H}=-\frac{\hbar^2\Delta}{2m}+V({\bf\hat{x}}).$$ (67)

Here, we have used the definition of the Laplace operator $\Delta=\nabla \cdot \nabla$. In Cartesian coordinates, it is $\Delta=\partial^2_x+\partial^2_y+\partial^2_z$. The Hamilton operator represents the total energy of the particle with mass $m$ in the potential $V({\bf x})$. We have introduced the hat as a notation for operators, but often the hat is omitted for simplicity. We make the important observation that $\hat{H}$ is exactly the expression that appears on the right hand side of the Schrödinger equation (1.21). This means we can write the Schrödinger equation as

$$i\hbar\frac{\partial}{\partial t}\Psi({\bf x},t)=\hat{ H}\Psi({\bf x},t).$$ (68)

This is the most general form of the Schrödinger equation in quantum mechanics. The replacement of ${\bf x}$ and ${\bf p}$ in quantum mechanics is

Axiom 3: The position ${\bf x}$ and momentum ${\bf p}$ are operators acting on wave functions,

$${\bf x} \to {\bf\hat{x}},\quad {\bf p} \to \frac{\hbar}{i}{\bf\nabla}.$$ (69)