1.4 Expectation Values

Generally speaking, if we make a measurement of a quantum mechanical system, the state of the system will be changed. This means that there is no additional precision to be gained from repeated measurements, unlike with classical systems. Once a measurement is made, the system’s wavefunction collapses into one of the eigenstates of the operator. The eigenvalues represent all possible measurement outcomes of that operator.

In order to make repeated measurements in the same way we would with a classical system, what we need to do is prepare a large number of systems in identical initial states, then measure them all. This gives us an ensemble average for the quantity, which is called the expectation value. Expectation values are denoted with angle brackets around the quantity, sometimes with a subscript $\psi$ to denote for which state we are taking the average value, e.g. $\langle\hat{x}\rangle_{\psi}$ represents the average position in the state $\psi$ .

Note that the name “expectation value” can be slightly misleading, for it does not denote what we “expect” to measure when we do a measurement. It simply means the average of many measurements of systems in identical states.

1.4.1 Operators with Discrete Eigenvalues

Suppose an operator $\hat{O}$ has a discrete set of measurement outcomes (eigenvalues) labelled $O_{n}$ . Then the expectation value is simply the sum of each outcome weighted by the probability:

\langle\hat{O}\rangle=\sum_{i}O_{i}P(O_{i}).

(1.29)

1.4.2 Operators with Continuous Eigenvalues

In the case that the set of measurement outcomes is continuous, the sum should change into an integral. But what should the integral look like?

We know that the wavefunction represents probability density in space, so the average value of position, which is simply the expectation value of the $\hat{x}$ operator, is

\langle\hat{x}(t)\rangle=\int_{-\infty}^{\infty}\hat{x}\psi(x,t)^{\ast}\psi(x,% t)\operatorname{\mathrm{d}}\!x=\int_{-\infty}^{\infty}x\lvert\psi(x,t)\rvert^{% 2}\operatorname{\mathrm{d}}\!x.

(1.30)

Notice that the time-dependence of $\langle\hat{x}\rangle$ comes purely from the wavefunction.

This is the quantum analogue of the trajectory in classical mechanics.

Can we extend this to other operators? For position, since $\hat{x}=x$ , it does not matter where we put $\hat{x}$ in the integral in equation 1.30, but for other operators, like momentum, it does matter. To see why, note that momentum contains a derivative $\frac{\partial}{\partial x}$ , then note that

\psi^{\ast}\frac{\partial\psi}{\partial x}\neq\frac{\partial\psi^{\ast}}{% \partial x}\psi,

(1.31)

in general.

Note that in classical mechanics, the trajectory follows Newton’s second law

\frac{\operatorname{\mathrm{d}}\!x(t)}{\operatorname{\mathrm{d}}\!t}=\frac{p(t% )}{m}.

(1.32)

We would like that the same thing would happen in classical mechanics, i.e. that

\frac{\operatorname{\mathrm{d}}\!\langle x(t)\rangle}{\operatorname{\mathrm{d}% }\!t}=\frac{\langle p(t)\rangle}{m},

(1.33)

but we do not know a priori if this is the case.

Let’s calculate the time derivative of $\langle x(t)\rangle$ to see what happens.

	$\displaystyle\frac{\operatorname{\mathrm{d}}\!\langle x(t)\rangle}{% \operatorname{\mathrm{d}}\!t}$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!t}% \left(\int_{-\infty}^{\infty}x\lvert\psi(x,t)\rvert^{2}\operatorname{\mathrm{d% }}\!x\right)$		(1.34)
		$\displaystyle=\int_{-\infty}^{\infty}x\left(\frac{\partial\psi^{\ast}}{% \partial t}\psi+\psi^{\ast}\frac{\partial\psi}{\partial t}\right)\operatorname% {\mathrm{d}}\!x.$		(1.35)

Now, we can change the temporal derivatives to spatial derivatives by rearranging the TDSE (equation 1.9) and its complex conjugate for $\frac{\partial\psi}{\partial t}$ and $\frac{\partial\psi^{\ast}}{\partial t}$ :

\frac{\partial\psi}{\partial t}=-\frac{\hbar}{2mi}\frac{\partial^{2}\psi}{% \partial x^{2}}+\frac{V}{i\hbar}\psi,\quad\frac{\partial\psi^{\ast}}{\partial t% }=\frac{\hbar}{2mi}\frac{\partial^{2}\psi^{\ast}}{\partial x^{2}}-\frac{V}{i% \hbar}\psi^{\ast}.

(1.36)

Substituting these in, we get

	$\displaystyle\frac{\operatorname{\mathrm{d}}\!\langle x(t)\rangle}{% \operatorname{\mathrm{d}}\!t}$	$\displaystyle=\frac{\hbar}{2mi}\int_{-\infty}^{\infty}x\left(\frac{\partial^{2% }\psi^{\ast}}{\partial x^{2}}\psi-\frac{V}{i\hbar}\lvert\psi\rvert-\psi^{\ast}% \frac{\partial^{2}\psi}{\partial x^{2}}+\frac{V}{i\hbar}\lvert\psi\rvert\right% )\operatorname{\mathrm{d}}\!x$		(1.37)
		$\displaystyle=\frac{\hbar}{2mi}\int_{-\infty}^{\infty}\left(\frac{\partial^{2}% \psi^{\ast}}{\partial x^{2}}x\psi-\psi^{\ast}x\frac{\partial^{2}\psi}{\partial x% ^{2}}\right)\operatorname{\mathrm{d}}\!x.$		(1.38)

Looking at the first term and integrating by parts, we get

	$\displaystyle\int_{-\infty}^{\infty}\frac{\partial^{2}\psi^{\ast}}{\partial x^% {2}}x\psi\operatorname{\mathrm{d}}\!x$	$\displaystyle=\left.\frac{\partial\psi^{\ast}}{\partial x}x\psi\right\|_{-% \infty}^{\infty}-\int_{-\infty}^{\infty}\frac{\partial\psi}{\partial x}\frac{% \partial}{\partial}(x\psi)\operatorname{\mathrm{d}}\!x$		(1.39)
		$\displaystyle=-\int_{-\infty}^{\infty}\frac{\partial\psi^{\ast}}{\partial x}% \left(\psi+x\frac{\partial\psi}{\partial x}\right)\operatorname{\mathrm{d}}\!x.$		(1.40)

The “surface” terms go to zero because the wavefunction is square integrable, and therefore must go to zero at $\pm\infty$ . Integrating by parts again and using the same trick, we get

	$\displaystyle\int_{-\infty}^{\infty}\frac{\partial^{2}\psi^{\ast}}{\partial x^% {2}}x\psi\operatorname{\mathrm{d}}\!x$	$\displaystyle=-\left.\psi^{\ast}\left(\psi+x\frac{\partial\psi}{\partial x}% \right)\right\|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\psi^{\ast}\left(2% \frac{\partial\psi}{\partial x}+x\frac{\partial^{2}\psi}{\partial x^{2}}\right% )\operatorname{\mathrm{d}}\!x$		(1.41)
		$\displaystyle=\int_{-\infty}^{\infty}\psi^{\ast}\left(2\frac{\partial\psi}{% \partial x}+x\frac{\partial^{2}\psi}{\partial x^{2}}\right)\operatorname{% \mathrm{d}}\!x.$		(1.42)

Substituting this into the time derivative for the expectation value for position, we get

$\displaystyle\frac{\operatorname{\mathrm{d}}\!\langle x(t)\rangle}{% \operatorname{\mathrm{d}}\!t}$	$\displaystyle=\frac{\hbar}{2mi}\int_{-\infty}^{\infty}\left(2\psi^{\ast}\frac{% \partial\psi}{\partial x}+\psi^{\ast}x\frac{\partial^{2}\psi}{\partial x^{2}}-% \psi^{\ast}x\frac{\partial^{2}\psi}{\partial x^{2}}\right)\operatorname{% \mathrm{d}}\!x$	(1.43)
	$\displaystyle=\frac{1}{m}\int_{-\infty}^{\infty}\psi^{\ast}\left(-i\hbar\frac{% \partial}{\partial x}\right)\psi\operatorname{\mathrm{d}}\!x$	(1.44)
	$\displaystyle=\frac{1}{m}\int_{-\infty}^{\infty}\psi^{\ast}\hat{p}\psi% \operatorname{\mathrm{d}}\!x,$	(1.45)

where we have substituted the definition of the momentum operator. So if we interpret this integral as the expectation value of momentum, equation 1.33 holds.

We thus have the following generalisation.

Definition 1.1.

For an operator $\hat{O}$ with continuous measurement outcomes, the expectation value of $\hat{O}$ for a state $\psi$ is defined as

\langle\hat{O}\rangle_{\psi}=\int_{-\infty}^{\infty}\psi(x,t)^{\ast}\hat{O}% \psi(x,t)\operatorname{\mathrm{d}}\!x.

(1.46)

In Dirac notation, this is denoted

\langle\hat{O}\rangle_{\psi}=\langle\psi\mathclose{}|\mathopen{}\hat{O}% \mathclose{}|\mathopen{}\psi\rangle.

(1.47)