# Expectation value meaning in quantum mechanics

While reading Griffiths' Introduction to Quantum Mechanics, I met the expected value

$$\langle x \rangle = \int \varPsi^\ast(x )\varPsi (x) \,\mathrm d x$$

and I have got some questions on this. The book says that

$$\langle p \rangle = \int \varPsi^\ast \left(\frac{\hbar} i\frac{\partial}{\partial x} \right) \varPsi (x) \,\mathrm d x$$

and from thease equations comes

$$\langle Q(x,p)\rangle =\int \varPsi^\ast Q \left( x,\frac{\hbar}i\frac{\partial}{\partial x} \right) \varPsi (x) \, \mathrm d x$$

Here $\langle Q(x,p)\rangle$ means "the expectation values of such a quantity", where such refers to the first two equations. But I can't understand, what expectation value is:

1. Is it energy? What will we get if the particle's momentum is $p$ and the position is $x$? Or does it mean something widely different?

2. I couldn't find any information in the book about $Q$ inside the integral. What doest that mean?

• Best to use \langle and \rangle for brackets rather than < and >, they look much nicer. – enumaris May 24 '18 at 20:02
• Note that in quantum mechanics, it is no longer meaningful to say "a particle with momentum $p$ and position $x$." We can only talk about states of a system, and states which correspond to unique values of both $p$ and $x$ do not exist. – J. Murray May 24 '18 at 20:17

## 4 Answers

The $Q$ here is simply a placeholder for any operator that is a function of $x$ and $p$. For example, if you want to get the expectation of the energy of a harmonic oscillator you would do: $$Q(x,p)\doteq E(x,p)=\frac{p^2}{2m}+\frac{1}{2}kx^2$$ $$\langle E(x,p)\rangle = \frac{1}{2m}\langle p^2\rangle+\frac{k}{2}\langle x^2\rangle = \frac{-\hbar^2}{2m}\int \Psi^*\left( \frac{\partial^2}{\partial^2 x}\right)\Psi dx+\frac{k}{2}\int\Psi^* (x^2)\Psi dx$$

The expectation value of any quantity is related to the average value from statistics and probability theory. In quantum mechanics the wave function is interpreted as the amplitude of a probability distribution (Copenhagen interpretation in place, but there have been others). Griffith's is abiding by this convention. The probability density is Pr = conjugate(Psi)*Psi, a real valued function. So, for example E(x) = integral(Pr * x), or mean(x) relative to the probability function properly normalized. In general, for an operator Op that corresponds to an observable, E(Op) = integral(conj(Psi) * Op(Psi)). The operator acts on the wave function then the result projected into the conjugate wave function. This provides a measure of the probability to observe a value in this state. The expectation value is interpreted as the statistical mean value you would get if you set up a large number of identically prepared experiments to measure the quantity Op. In Griffith's notation Q(x,p) means that the observable quantity Q can be a function of position and momentum. Energy, or the Hamiltonian, is the most famous one as its states are "stationary". But the wave function appearing in the equation need not be an eigenstate of the Hamiltonian.

The expectation value of $Q$ is the result of a measurement of $Q$ that QM predicts for a system in the state $\psi$. Common examples of $Q$ are the total energy, the kinetic energy $p^2/2m$, the potential energy $V(x)$ or the angular momentum $\vec r \times \vec p$.

The expectation value of $\langle \psi |\hat Q|\psi\rangle$ is the average result of a measurement of $Q$ for the state $\psi$. If $\psi$ is an eigenstate of $\hat Q$ it is the eigenvalue. This is usually the case if $\hat Q$ is the energy or the angular momentum. If it is not then the expectation value is the sum of (or integral over) the probability of each eigenstate times its eigenvalue. This is usually the case if $\hat Q$ is kinetic or potential energy. Also position or momentum.

It's worth emphasising that the expectation value does not have to be a possible value. The $z$ component of an electron spin can be $+{1 \over 2} \hbar$ or $-{1\over 2} \hbar$. The expectation value for an electron previously polarised along the $x$ direction is $0$ as there is an equal probability of + or -, even though $0$ is not a possible result