# Classical versus quantum momentum

1. Jun 4, 2015

### DiracPool

I have a question that's been gnawing at me recently. In classical physics, momentum is mass times velocity, which makes sense, take the mass of an object, calculate the velocity and you are left with a vector quantity of momentum.

In quantum physics, we have momentum p=h/λ. This doesn't seem to be a vector quantity and the mass and the velocity are both simply represented by a wavelength? Since this DeBroglie formula represents actual massive particles, how do we unpack a wavelength to find out the actual mass and velocity of the particle (say electron), as a vector quantity?

2. Jun 4, 2015

### bhobba

That's not momentum in QM. Its an operator:
http://en.wikipedia.org/wiki/Momentum_operator

In fact the underlying reason for momentum, and why it is that, is exactly the same in QM and Classical Mechanics - its related to translational symmetry of a system.

However that is a highly advanced view you need considerable background to come to grips with. At the beginner level you have to accept that's the way it is.

But as a reference from QM see Chapter 3 Of Ballentine, and for Classical Mechanics see Landau or pretty much any book on advanced mechanics:
http://www.eftaylor.com/pub/HancManuscriptSymmetry.pdf

Thanks
Bill

3. Jun 4, 2015

### DiracPool

I thought the quantum mechanical operator was , and that it had to operate on a wavefunction to yield an eigenvalue, which still doesn't unpack my above concern. p=h/λ is a stand alone equation assigning a momentum value to any given particle, isn't it? I understand the concept of translational symmetry and the conservation of momentum and time translation symmetry and the conservation of energy, but I don't see how we unpack a mass and directional velocity out of p=h/λ, or even using the operator on a wave function.

4. Jun 4, 2015

### bhobba

It isn't.

You are getting confused in what QM says.

Try applying it, for example, to the energy eigenstates of an electron in a hydrogen atom:
http://www.ciul.ul.pt/~ananunes/QM/Laguerres&Hydrogenatom.pdf [Broken]

What's the momentum of the 1S state?

Thanks
Bill

Last edited by a moderator: May 7, 2017
5. Jun 4, 2015

### vanhees71

First of all momentum in quantum theory is generally not $m v$. Its rather associated with canonical momentum than mechanical momentum, but that's a detail (important e.g. in the case of a charged particle in a magnetic field).

The operator $\hat{p}=-\mathrm{i} \vec{\nabla}$ (setting $\hbar=1$ for convenience) acts on wave functions $\psi(t,x)$. The wave function needs not be a momentum eigenvector. It's also easy to see that the momentum eigenvectors are generalized functions and no wave functions, because they are not square integrable. Indeed, the eigenvector equation reads
$$-\mathrm{i} \vec{\nabla} u_{\vec{p}}(\vec{x})=\vec{p} u_{\vec{p}}(\vec{x}) \; \Rightarrow \; u_{\vec{p}}(\vec{x})=N \exp(\mathrm{i} \vec{p} \cdot \vec{x}), \quad \vec{p} \in \mathbb{R}^3.$$
The normalization constant $N$ is usually chosen such that the eigenvectors fulfill the generalized orthonormality condition
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}'}^*(\vec{x}) u_{\vec{p}}(\vec{x})=\delta^{(3)}(\vec{p}-\vec{p}') \; \Rightarrow \; N=\frac{1}{\sqrt{(2\pi)^3}}.$$
The wave length of such a plane-wave momentum eigenfunction is then of course $\lambda=2 \pi/p$.

I do not understand your question concerning "unpacking a mass and directional velocity". Perhaps what you mean is to ask about solutions of the time-dependent Schrödinger equation for free particles and how to figure out their average velocity given such a state?

$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}=-\frac{1}{2m} \vec{\nabla}^2.$$
Now given an arbitrary normalizable state at $t=0$, $|\psi_0 \rangle$ the solution of the corresponding time-dependent Schrödinger equation is easily expressed in terms of momentum eigenfunctions:
$$\psi(t,\vec{x})=\langle \vec{x}|-\mathrm{i} \hat{H} t \psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \langle \vec{x}|\vec{p} \rangle \langle \vec{p}|\exp(-\mathrm{i} \hat{H})\psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} u_{\vec{p}}(\vec{x}) \exp \left (-\mathrm{i} \frac{\vec{p}^2}{2m} \right ) \tilde{\psi}_0(\vec{p}),$$
where
$$\tilde{\psi}_0(\vec{p})=\langle \vec{p}|\psi_0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \langle \vec{p}|\vec{x} \rangle \langle \vec{x} |\psi_0 \rangle = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{1}{\sqrt{(2 \pi)^3}} \exp(-\mathrm{i} \vec{p} \cdot \vec{x}) \psi_0(\vec{x}).$$
As you see, the transformation between the position and momentum representation of the Hilbert-space vectors (wave functions in the position and momentum representation) are the Fourier transformations.

Now the operator for the time derivative of the observable is given by the commutator with the Hamiltonian, i.e., in your case of a free particle
$\hat{\vec{v}}=\frac{1}{\mathrm{i}} [\hat{\vec{x}},\hat{H}]=\frac{1}{m} \hat{\vec{p}}=-\frac{\mathrm{i}}{m} \vec{\nabla}.$
In this case you have indeed the usual relation between velocity and momentum. So you get your expectation value for the velocity by
$$\langle \vec{v} \rangle=-\frac{\mathrm{i}}{m} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi^*(t,\vec{x}) \vec{\nabla} \psi(t,\vec{x}),$$
where I've assumed that the wave function is properly normalized,
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \psi^*(t,\vec{x}) \psi(t,\vec{x}) = 1.$$

6. Jun 4, 2015

### bhobba

That's true.

But what I was hoping to get across to the OP was that most of the time, as explained in Chapter 3 of Ballentine, where V is the velocity operator defined in that chapter, momentum is exactly the same as classical mechanics ie mV. V is a vector operator ie is 3 operators corresponding to each direction in whatever coordinate system we are using - which of course you also explained.

Thanks
Bill

Last edited: Jun 4, 2015
7. Jun 4, 2015

### DiracPool

Thanks vanhees, I was looking for a bit more of a colloquial discussion on the matter, but what I'm seeing here is that the reconciliation is not so straightforward. I think I'll start by reviewing the Ballentine reference, thanks Bhobba.

8. Jun 4, 2015

### Staff: Mentor

That's the operator that gives you the $x$ component of the momentum. There are similar operators for the $y$ and $z$ components. In a three-dimensional problem you'll write the wave function as $\psi(x,y,z,t)$ instead of just $\psi(x,t)$, you'll need all three operators, and the "vectorness" of the momentum will be back.

Intro texts usually focus on one-dimensional problems because they are the simplest way to get the essential points of position and momentum across, and that's what you're seeing here. In a one dimensional problem there are only two directions, positive and negative.

9. Jun 4, 2015

### DiracPool

Ok, that's what I was looking for. So you can recover the velocity of the particle by applying the operator on all 3 axes? Do you need the position operator and the energy operator too? I guess I'm intrigued with this idea that "all the information that can possibly be known about a system (particle) is contained in the wave equation." When I mentioned "unpacking" the wavelength in terms of translating that to an actual particle with a certain mass traveling in a certain direction, that's kind of what I was getting at. How do you do that?

I think you do need the 3 axes. I was watching this video on Brightstorm where he related a mass and a velocity to the wavelength, but it seemed to be a scalar value that didn't give much information about the dynamics of the system. Fast forward to 2:45..

10. Jun 4, 2015

### bhobba

No.

The velocity operator is like any other operator in QM:
http://www.tcm.phy.cam.ac.uk/~bds10/aqp/handout_operator.pdf

Here are the two axioms of QM:
Axiom 1
Associated with each Von Neumann measurement we can find a Hermitian operator O, called the observations observable such that the possible outcomes of the observation are its eigenvalues yi.

Axiiom 2 - called the Born Rule
Associated with any system is a positive operator of unit trace, P, called the state of the system, such that expected value of of the outcomes of the observation is Trace (PO).

Thanks
Bill

11. Jun 4, 2015

### Staff: Mentor

Kind of sort of yes, but only kind of sort of... Typically the particle won't be in an eigenstate of the momentum operator, so the quantity $<p>/m$ which you're trying to call the velocity is just going to be an expectation value itself. It's also a remarkably unhelpful one because if you're working with objects large enough to maintain a reasonably predictable velocity over multiple measurements, you're working with objects large enough and slow enough that classical mechanics will work just fine - this is why you don't find a lot of velocity problems in the exercises section of a QM textbook.

The momentum is nearly always a more useful and physically relevant quantity (especially because it is conserved!).

Also I'd like to point out something that you may have missed in VanHees71's post - he used the general definition oif the momentum operator in terms of $\vec{\nabla}$, a definition that wraps the derivatives with respect to all three dimensions into a single vectorish thing.