I know the equation for the momentum operator, but what exactly is the momentum operator? It's bizarre to me that taking the derivative of the wave function, which is an operator, should return something that isn't just a function. Specifically, my confusion is that first momentum is written as being multiplied by position, but then, after the derivative, it's acting on a function. I feel like there is some formalism missing. How does the operator go from being multiplied to acting on something?


2 Answers 2


The wave function isn't an operator; the word "operator" in quantum mechanics means something more precise than "function". You might say that the momentum operator is "something to put in an integral to get the expectation value of momentum"; let me explain.

We know that for a particle in a state $\psi$,

$$ \langle x \rangle = \int_{-\infty}^{+\infty} x |\psi(x,t)|^2 \mathrm{d}x $$

because $|\psi(x,t)|^2\mathrm{d}x$ is the probability that the particle will be found in the small interval $(x,x+\mathrm{d}x)$ at time $t$. Let's differentiate this and push the derivative into the integral

$$ \frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = \int_{-\infty}^{+\infty} x \frac{\partial|\psi(x,t)|^2}{\partial t} \mathrm{d}x $$

Now we use Schrodinger's equation to replace the time derivative with space derivatives

$$ \frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = \frac{i\hbar}{2m}\int x \frac{\partial}{\partial x} \left(\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial \psi^*}{\partial x}\right) \mathrm{d}x $$

Use integration by parts to eliminate integrate away the outer $\frac{\partial}{\partial x}$ and differentiate the $x$

$$ \frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = -\frac{i\hbar}{2m}\int \left(\psi^*\frac{\partial\psi}{\partial x} - \psi\frac{\partial \psi^*}{\partial x}\right) \mathrm{d}x $$

Perform another integration by parts on the second term

$$ \frac{\mathrm{d} \langle x \rangle}{\mathrm{d}t} = -\frac{i\hbar}{m}\int \left(\psi^*\frac{\partial\psi}{\partial x}\right) \mathrm{d}x $$

Multiply by $m$

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

Notice that this is the same form as the integral for $\langle x \rangle$, except we use $\frac{\hbar}{i}\frac{\partial}{\partial x}$ in place of $x$. That's why they're called the momentum and position operators respectively; they are the operators you place between $\psi^*$ and $\psi$ in the integral to obtain the expectation value of that variable. There are also operators for angular momentum, energy, etc.

Of course, there are other more useful definitions of an operator. For example, notice that if we write $\psi$ as a sum of sinusoidal functions, each sinusoidal function is an eigenfunction of the momentum operator, and the integral becomes very simple to evaluate; hence we could think of the momentum operator as an operator that returns the sum of the component sinusoidal functions (momentum eigenstates) of $\psi$, weighted by momentum, which is how I like to think of it as.


The momentum operator, like other operators in quantum mechanics, acts on a given wave function (state). Multiplication of operators in linear algebra is the same as them "acting on" a mathematical object. Originally quantum mechanics was called 'matrix mechanics'; so when you study linear algebra you're really studying quantum mechanics too. Momentum is a Hermetian operator, and it's eigenvalues correspond to the possible values the momentum can take on in a given measurement. When the momentum operator $ \mathbf{A} $ "acts on" a given state $ \langle a | $ ("state" here is equivalent to "eigenvector"), the state has a corresponding eigenvalue $ a $. We would write this in an equation as $$ \mathbf{A} \langle a | = a \langle a | $$ This means that in a measurement of the observable $\mathbf{A}$ of a system in a state $ \langle a | $ you would return a value of $a$ within limits of uncertainty.

  • $\begingroup$ I still don't understand how momentum went from being in the exponent and being multiplied by a variable, to being in front of the wave function, and acting on it. $\endgroup$
    – user24082
    Commented Sep 15, 2013 at 23:04
  • $\begingroup$ Taking the derivative of $\psi$ is the same as multiplying $\psi$ by $\frac{i}{h}\mathbf{p}$ (definition of the derivative of an exponential). From linear algebra, taking derivatives is the same as a matrix ("operator") multiplying the function you want to derive. It follows that for the wave function, momentum equals $-ih \frac{\partial}{\partial x} $ (or $-ih\nabla$ in 3D). $\endgroup$
    – rchwn
    Commented Sep 15, 2013 at 23:16