What's the meaning of the momentum operator?

Question

I understand that a wave function $\psi(x, t)$ tells me that the probability to find the particle at position $x$ is $|\psi(x, t)|^2$. In the Schrodinger equation, we use the momentum operator $\hat{p} = -i \hbar \frac{\partial}{\partial x}$, but I'm having trouble understanding what the result $\hat{p}\psi$ means.

Is it still called a wave function? If so, what is the meaning of the value $\hat{p}\psi(x, t)$ for a given $x$? If $|\hat{p}\psi(x, t)|^2$ still represents a probability, what is it a probability of exactly?

The reason I'm asking is that I've just started reading about quantum mechanics and the Schrodinger equation is explained as $\hat{E}\psi = \frac{\hat{p} \cdot \hat{p}}{2m} + \hat{V}\psi$ in an analogy to classical mechanics, but I have no idea why the energy and momentum operators are what they are, or how to even interpret them momentum or energy being a function of position. I feel like I'm missing something basic.

Answer 1

Disperson relation

Before even looking at the Schrödinger eqaution let's look at the wave equation to understand an important concept in physics: the dispersion relation. The wave equation (1D) is given by $$\frac{1}{c^2}\frac{\partial^2u}{\partial t^2}-\frac{\partial^2u}{\partial x^2}=0.$$ One way to find solutions of this equation is to go to Fourier space. This means that I look at $u$ as a function of frequency, instead of space or time. In Fourier space, taking the derivative of a function becomes multiplication by frequency. I define the Fourier transform as \begin{align} u(t, x)=\frac{1}{(2\pi)^2}\int\mathrm dk\,\mathrm d\omega\,\hat u(\omega,k)e^{i kx}e^{i\omega t}. \end{align} Here $k$ is called the angular wavenumber and $\omega$ the angular frequency. They are related to wavelength and frequency by \begin{align} k=\frac{2\pi}{\lambda}\\ \omega=2\pi f \end{align} If we now plug this into the wave equation, we get after some work, \begin{align} \frac{1}{(2\pi)^2}\int\mathrm dk\,\mathrm d\omega\,\left[-\frac{\omega^2}{c^2}+k^2\right]\hat u(\omega,k)e^{i kx}e^{i\omega t}=0. \end{align} This must hold true for all functions $\hat u$, so for each point $(\omega,k)$ either $\hat u=0$ or $\omega^2=c^2k^2$. This means that all non-zero values of $\hat u$ occur along the two lines $\omega=\pm ck$. The function $\omega^2=c^2k^2$ is called the dispersion relation. For valid solutions of the wave equation, it relates frequency to wavelength.

The Schrödinger equation

(I do not know a lot about the history, so excuse me if I butcher this section and please correct me if necessary)

In the early 1900's, Einstein and Planck postulated that light came in discrete packets, called photons. This explained the long standing mystery of the ultraviolet catastrophe, as well as the photo-electric effect. The energy of each photon is related to its frequency, by the simple relation $$E=hf.$$ Sometime later, De Broglie postulated that matter was made out of waves. He did this to explain the discrete nature of absorption spectra of atoms. He used the following relation for momentum:

$$p=\frac{h}{\lambda}$$

Now, if matter is made out of waves, surely there must be some equation that describes the evolution of this wave. Schrödinger was the first to find a satisfactory one. Let's pretend we are Schrödinger and "derive" this equation.

According to classical physics, matter has the following expression for energy (Hamiltonian) $$E=\frac{p^2}{2m}+U(x)$$ Now, let's use $E=hf$ and $p=h/\lambda$. I first write them as $E=\hbar \omega$ and $p=\hbar k$. $$\hbar \omega=\frac{\hbar^2 k^2}{2m}+\hat U(k)$$ I replaced $U(x)\mapsto \hat U(k)$. This will become clear soon. We want to derive a wave-like equation, so let's interpret this equation as a dispersion relation and work in reverse.

\begin{align} \frac{1}{(2\pi)^2}\int\mathrm dk\,\mathrm d\omega\,\left[ \hbar\omega - \frac{\hbar^2 k^2}{2m}-\hat U(k) \right]\hat \psi(\omega,k)e^{i kx}e^{i\omega t}&=0\\ \left[ i\hbar\frac{\partial}{\partial t}+ \frac{\hbar^2 }{2m}\frac{\partial^2}{\partial x^2}-U(x) \right]\frac{1}{(2\pi)^2}\int\mathrm dk\,\mathrm d\omega\,\hat \psi(\omega,k)e^{i kx}e^{i\omega t}&=0\\ \left[ i\hbar\frac{\partial}{\partial t}+ \frac{\hbar^2 }{2m}\frac{\partial^2}{\partial x^2}-U(x) \right]\psi(x)&=0 \end{align} and finally we have "derived" the Schrödinger equation. This derivation is much easier when you know how the final answer looks, so don't underestimate how hard this was to find back in the days.

What does $\hat p=-i\frac{\partial}{\partial x}$ mean?

So far, identifying derivatives with $p$ has just been a tool to get the right dispersion relation, but how do we interpret this? For a plane wave this is easy to interpret. For a plane wave $e^{ikx/\hbar}$ we recognize the momentum $p=\hbar k$. The momentum operator acts like $\hat p{e^{ikx/\hbar}}=(\hbar k){e^{ikx/\hbar}}=p{e^{ikx/\hbar}}$. In other words, it multiplies by the momentum. The new function $\hat p\psi(x) $ is in itself meaningless, the same way that $x\psi(x)$ is meaningless. But, since the Schrödinger equation is linear, superpositions of solutions are also solutions to the Schrödinger equation. So we can interpret a superposition of two plane waves with definite momentum as having "two momenta at the same time". In other words,

$$|\psi\rangle=c_1|p_1\rangle+c_2|p_2\rangle,$$ respresents a state that has both $p_1$ as momentum and $p_2$ as momentum. By the Born rule, we say that the average momentum is $|c_1|^2p_1+|c_2|^2p_2$. We can calculate this using $\langle\psi|\hat p|\psi\rangle$. When $\hat p$ acts on $|\psi\rangle$, it multiplies every eigenstate with its momentum. The multiplication by $\langle \psi|$ then turns that expression into an expectation value.

The fact that $\langle\psi|\hat p|\psi\rangle$ is the expectation value of $p$ relies on two facts

1. $\hat p$ multiplies each state $|p_i\rangle$ with $p_i$
2. The states $|p_i\rangle$ form an orthonormal basis. This means that $$\langle \psi_i|\psi_j\rangle=\cases{1&i=j\\0&i$\neq j$}$$

It turns out that both these conditions are fulfilled if $\hat p$ is Hermitian, which implies $\hat p^\dagger=\hat p$. This also guarantees real eigenvalues, so the expectation value is real. This means that we can interpret every Hermitian operator as an observable.

Finally

Generally speaking, $\hat O\psi(x)$ for an observable $\hat O$ does not have a nice interpretation, that's just not now how quantum mechanics works. What we can say though, is that the eigenstates of $\hat O$ represent all possible states of $\hat O$, the eigenvalues of $\hat O$ represent all possible values of $\hat O$ and $\psi|\hat O|\psi\rangle$ represents the expected outcome of a measurement of $\hat O$.

Reference: https://www.mpiwg-berlin.mpg.de/Preprints/P437.PDF

Answer 2

Quantum mechanics uses notions of systems, states, observables, and dynamics. These notions pervade all textbook physical theories.

A system is any object we want to study the properties of. We say that a system occupies a state, which, operationally, completely specifies information about the system (as much as the physical theory allows). An observable is (usually) a physical quantity of interest. Given the system in some initial state, a dynamic relation (usually a differential equation) tells us how the state changes as time evolves.

In (pure) quantum mechanics, systems are Hilbert spaces, or complex vector spaces. We denote such structures by $\mathcal{H}$. States of the system are unit vectors of the space, denoted $\lvert \psi \rangle \in \mathcal{H}$. Observables are Hermitian operators over Hilbert space $$\hat{O}: \mathcal{H}\to \mathcal{H} \text{ such that } \hat{O}^\dagger = \hat{O}.$$ The dynamic relation is the Schrodinger equation $$i\hbar \frac{\partial}{\partial t} \lvert \psi\rangle = \hat{H}\lvert \psi \rangle.$$ Finally, we must posit a rule that connects all these mathematics to something in the real world. This is called Born's rule $$\text{Pr}(\lvert \psi \rangle \to \lvert a \rangle)= \lvert \langle a \lvert \psi \rangle \lvert^2,$$ which gives the probability of obtaining the measurement outcome $a$ having made a measurement of the observable $\hat{A}$ on initial state $\lvert \psi \rangle$. I have stated these facts in a basis independent manner. Almost all of the quantities in your question are written in the position basis, which may be contributing to your confusion.

Now, to get to your question. In classical mechanics, observables always take on definite values. For example, I can solve for the trajectory of a point-like object using Newtonian mechanics. In doing so, I will obtain equations that exactly describe the position of the point-like object $\{x(t), y(t)\}$.

In quantum mechanics, observables do not always take on definite values. In particular, we are forced to consider the expectation values (an average) of observables with respect to a state of the system. Expectation values are computed as $$\langle \psi \lvert \hat{O} \lvert \psi \rangle.$$

Now, let $\hat{p}$ be the momentum operator. The quantity $\hat{p}\lvert \psi \rangle$ is indeed another state (recall that observables are simply maps over the space of states). It, to my understanding, doesn't really have a generally useful physical interpretation.