I've recently started an introductory course in Quantum Mechanics and I'm having some trouble understanding what the expectation of an operator is. I understand how we get the formula for the expectation of position, if we assume that the complex function $\psi$ describes a particle, and $f=|\psi|^2$ is a probability density of finding the particle at varying x at time t.
$\langle x\rangle=\int^{\infty}_{-\infty} xf(x)dx=\int^{\infty}_{-\infty}x\psi^*(x) \psi(x)dx=\int^{\infty}_{-\infty}\psi^*(x)x \psi(x)dx$
But in my course the expectation of an operator (for example momentum $\hat{p}=-i\bar{h}\frac{\partial}{\partial x}$) is defined as
$\langle\hat{p}\rangle=\int^{\infty}_{-\infty}\psi^*(x)\hat{p} \psi(x)dx=\int^{\infty}_{-\infty}\psi^*(x)-i\bar{h}\frac{\partial}{\partial x} \psi(x)dx=-i\bar{h}\int^{\infty}_{-\infty}\psi^*(x)\frac{\partial}{\partial x} \psi(x)$
I don't understand what the expectation of an operator would mean. I can understand the phrase "we expect the momentum to be $p_0$" to mean "after doing this experiment a large number of times, the mean momentum recorded will be $p_0$." But I can't understand what the expectation of a derivative is, before it's even applied to anything!
What does this expected value of the momentum operation mean in QM? Does it relate to something in the real world? Did we just notice that the formula above just happens to look like the average result after a lot of experiments?
Anyway, any clarification would be well appreciated!
Best Answer
The answer to your question begins by looking back at one of the oldest ideas in quantum mechanics, the deBroglie relation between the momentum of a particle and the wavelength of the associated wave. (This, along with Planck's relation between energy of particles and frequency of waves, was the basic connection between the particle and wave pictures, which underlies quantum theory.) The deBroglie relation reads $p=h/\lambda$, where $p$ is the particle's momentum and $\lambda$ is the wavelength of the associated wave. (This is a one-dimensional equation; for particles in three dimensions one also needs to take the direction of the momentum into account, but for simplicity I'll confine this answer to one dimension; the formulas in the question are also one-dimensional.) In particular, if the particle has a definite momentum $p$, then the corresponding wave is (a constant times) $\psi(x)=\exp(ipx/\hbar)$. Usually, a particle won't have a definite momentum (uncertainty principle), because its wave function $\psi$ will be a superposition of many waves of the form $\exp(ipx/\hbar)$ for different values of $p$, rather than a single such wave. But before getting into superpositions, let me point out that $$ -i\hbar\frac d{dx}\exp(ipx/\hbar)=p\exp(ipx/\hbar). $$ (In a suitable function space, this makes $\exp(ipx/\hbar)$ and eigenfunction of the operator $-i\hbar\frac d{dx}$, but that suitable space won't be the Hilbert space $L^2$, because $\exp(ipx/\hbar)$ isn't in that space; its norm is infinite; this leads to technicalities that are, fortunately, irrelevant here.) This equation is the fundamental reason why quantum mechanics regards the operator $-i\hbar\frac d{dx}$ as representing momentum.
Now let me look at superpositions. Suppose the wave function $\psi(x)$ is a superposition of functions $\exp(ipx/\hbar)$ for various valuse of $p$: $$ \psi(x)=\int_{-\infty}^{\infty}\hat\psi(p)\exp(ipx/\hbar)\,dp. $$ Mathematically, $\hat\psi$ here is the Fourier transform (or inverse Fourier transform, or, depending on your conventions, one of these times $2\pi$, or times $\sqrt{2\pi}$) of $\psi$. Physically (because mathematicians aren't supposed to say such things in public), $|\hat\psi(p)|^2\,dp$ is the probability of finding the particle to have momentum in an infinitesimal interval of length $dp$ at $p$. (This is exactly analogous to $|\psi(x)|^2\,dx$ being the probability of finding the particle in an infinitesimal interval of length $dx$ at position $x$.) With this physical interpretation of $\hat\psi$, we get that the expectation of the momentum is $$ \int_{-\infty}^\infty p|\hat\psi(p)|^2\,dp. $$ The integrand here is the product of $\hat\psi(p)^*$ and $p\hat\psi(p)$. So the integral is the inner product, in $L^2$, of $\hat\psi(p)$ and $p\hat\psi(p)$. The first of these is the Fourier transform of $\psi(x)$, and the second is, by the "eigenvalue" observation above, the Fourier transform of$-i\hbar\frac d{dx}\psi(x)$. Using the fact that the Fourier transform preserves inner products (if one has factors of $2\pi$ in the right places, which I won't worry about here), we get that the integral for the expectation of the momentum equals $$ \int_{-\infty}^\infty\psi(x)^*(-i\hbar\frac d{dx})\psi(x)\,dx. $$