# Momentum expected value derivation. From classical form to quantum operator

I found in many places that the average momentum of a particle is given by:

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

I think that it comes from considering the classical momentum:

$$\langle p\rangle=m\frac{\mathrm{d}\langle x\rangle}{\mathrm{d}t}$$

and that the expected value of the position is given by:

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

But when replacing $\langle x\rangle$ and differentiating inside the integral I don't know how to handle the derivatives of $\psi$ for getting the average momentum formula. Any suggestion?

• Your first equation is just $\langle\psi|\hat{p}|\psi\rangle$ where $\hat{p}$ is the momentum operator $\hat{p}=-i\hbar\frac{d}{dx}$. Jul 14, 2017 at 4:15
• What is ⟨ψ|p^|ψ⟩, I don't get the notation. Sorry. Jul 14, 2017 at 4:20

Since your question is about the single-particle momentum operator, I will assume that we are dealing with a single particle moving in an arbitrary scalar potential. Suppose we start with

$$\langle p \rangle = m\frac{d\langle x\rangle}{dt}$$

and

$$\langle x\rangle = \int x|\psi|^2 dx = \int \psi^* x \psi\; dx.$$

Applying the time derivative to the above yields

$$\frac{d\langle x\rangle}{dt}=\int\frac{\partial\psi^*}{\partial t}x\psi+\psi^*\frac{\partial x}{\partial t}\psi+\psi^*x\frac{\partial\psi}{\partial t}\;dx$$

Since the operator $x$ has no explicit time dependence, the middle term is zero. In addition, the Schrodinger equation states

$$\frac{\partial\psi}{\partial t} = \frac{1}{i\hbar}H\psi.$$

Applying this to $\psi^*$, we have

$$\frac{\partial\psi^*}{\partial t} = \frac{-1}{i\hbar}\psi^*H$$

and substituting both of these in for the appropriate quantities in $\frac{d\langle x\rangle}{dt}$:

$$\frac{d\langle x\rangle}{dt} = \int\frac{-1}{i\hbar}\psi^*Hx\psi+\frac{1}{i\hbar}\psi^*xH\psi\; dx=\frac{1}{i\hbar}\int\psi^*[x,H]\psi\; dx.$$

Now all that is left is to calculate the commutator $[x,H]$. Since we're dealing with a single particle in an arbitrary scalar potential, we can write $H=\frac{p^2}{2m}+V(x,t)$ so that:

$$[x,H]=\left[x,\frac{p^2}{2m}+V(x,t)\right]=\left[x,\frac{p^2}{2m}\right]+[x,V(x,t)].$$

Since $V(x,t)$ is a function of $x$ alone*, we have that $[x,V(x,t)]=0$. We then use the commutator identity

$$[A,BC]=[A,B]C+B[A,C]$$

to write

$$\left[x,\frac{p^2}{2m}\right] = \frac{1}{2m}([x,p]p+p[x,p])=\frac{1}{2m}(i\hbar p+pi\hbar)=\frac{i\hbar}{m}p$$

since $[x,p]=i\hbar$. The representation of $p$ in the position basis is $-i\hbar\frac{\partial}{\partial x}$, so in the position basis

$$[x,H]=\frac{\hbar^2}{m}\frac{\partial}{\partial x}.$$

Finally, substituting, we have that

$$\frac{d\langle x\rangle}{dt} = \frac{1}{i\hbar}\int\psi^*[x,H]\psi\; dx = \int \psi^*\left(\frac{\hbar}{im}\right)\frac{\partial\psi}{\partial x}\; dx$$

so that

$$\langle p\rangle = m\frac{d\langle x\rangle}{dt} = \int \psi^*\left(\frac{\hbar}{i}\right)\frac{\partial\psi}{\partial x}\; dx.$$

*What I mean here is that $V$ is not a function of any other operators, and $t$ isn't an operator in quantum mechanics, only a parameter.

• I want to know how to get the average momentum formula from the momentum definition. Jul 14, 2017 at 4:35
• @JoshuaSalazar From the definition of the momentum operator in position space? Or from the classical definition? What exactly do you want to start with? Jul 14, 2017 at 4:36
• The classical definition Jul 14, 2017 at 4:44
• @JoshuaSalazar Edited. Jul 14, 2017 at 5:49
• @JoshuaSalazar That's what happens when you take the complex conjugate of a product of operators. It's an identity that arises from group theory. The Hamiltonian is Hermitian, so its complex conjugate is itself. Jul 14, 2017 at 6:11

$m \frac{d}{dt} \langle x \rangle = m \frac{d}{dt} \int dx \ \Psi^* x \Psi$

Use product rule to get the above into the form:

$= m \int dx \left[\frac{\partial \Psi^*}{\partial t} x\Psi + \Psi^* \frac{\partial x}{\partial t} \Psi + \Psi^* x \frac{\partial \Psi}{\partial t} \right] \ \ \ \ \ -(1)$

The second term contains $\frac{\partial x}{\partial t}$, which is $0$.

This is just calculus. Now comes the crucial step of imposing physics: the Schrodinger equation: $i \hbar \frac{\partial \Psi}{\partial t} = \hat{H} \Psi$ (and also $-i \hbar \frac{\partial \Psi^*}{\partial t} = \hat{H} \Psi^*$)

Write the operator $\hat{H}$ in terms of second order spatial derivative (acting on $\Psi$ and $\Psi^*$). Through Schrodinger's equation, you get a relation between second order spatial derivative and first order time derivative. Replace the first order time derivatives in $(1)$ with second order spatial derivatives. And then integrate by parts, to reduce the second order spatial derivative to first order spatial derivative, using the boundary condition that $\frac{d\Psi}{dx}\left(\text{and} \frac{d\Psi^*}{dx}\right)$ go to zero at infinity. After a few steps of algebra, and you get it to the form $\int dx \ \Psi^* \left( -i\hbar \frac{\partial \Psi}{\partial x} \right)$, which is what you wanted: $\langle \hat{p} \rangle$.

This (or a similar) calculation is usually given in various resources. Can you take it from here and do it yourself?

• Now I get it more clearly. But, I suppose that the $V(x,t)$ terms would vanish at some point right? Or should I consider the potential to be zero? Jul 14, 2017 at 5:57
• Yes, they will cancel out. You don't have to assume $V=0$ Jul 14, 2017 at 6:13
• @Avantgarde : only if the potential is real will the relevant terms cancel. (Complex $Vs$ occur in simple models of unstable particles, and possibly elsewhere. The imaginary part of a complex $V$ functions like a damping term.) Jul 14, 2017 at 11:50
• @ZeroTheHero Yeah, I did not add that caveat. Thanks Jul 14, 2017 at 12:07