# Momentum an electron leads to velocity

1. Apr 29, 2006

### pivoxa15

The momentum of an electron is a sharp observable and the average of the momentum is 0. I assume this is because the velocity is direction dependent and positive, negative cancels. But for a particle in a 1D well, the absolute momentum is n(hbar)(pie)/L.

Momentum is mv. The mass of the electron is a positive constant so v is a positive constant which depends on the excitation state of the electron. Hence having a momentum would mean having a velocity. But there is no trajectory of an electron. How do we measure the velocity of something without a trajectory? Velocity is defined as delx/delt. Could it happen that two successive measurements of the electron leads to delx/delt > c? Theoretically, I could tweak the above momentum formula and make L around 10^-12m for v=c although as always in QM the HP might intervene somehow.

I suppose you could also define momentum as srt(2Em) or something else that does not depend on velocity. But I am curious about the idea of the velocity of an electron and whether it is possible to have one at all? And what would it mean for it to be a velocity?

Last edited: Apr 29, 2006
2. Apr 29, 2006

### Staff: Mentor

I assume you're talking about the infinite square well specifically. The momentum for a given state (value of n) is not sharp! When you measure it, you get either $+\sqrt{2ME}$ or $-\sqrt{2ME}$, at random. The expectation value $<p> = 0$, but $<p^2> \ne 0$ so $\Delta p \ne 0$.

Have you studied QM operators yet? If so, you'll recall that when we use the position wavefunction $\psi(x,t)$ to represent a particle, then its momentum is represented by the operator

$$\hat p = -i \hbar \frac {\partial}{\partial x}[/itex] I think the only sensible way to represent the velocity is as the operator [tex]\hat v = \frac {\hat p}{m} = -\frac{i \hbar}{m} \frac {\partial}{\partial x}[/itex] This obviously means that the expectation values are related as $<p> = m<v>$. I don't think it makes sense to define velocity as dx/dt because you can't measure the position twice in succession and get repeatable results. After you measure the position of a particle, it's not in the same state that it was in before. I need to go in a few minutes, so I have to break off here, but I'll think about this a bit more. In the meantime hopefully someone else will chip in. Last edited: Apr 29, 2006 3. Apr 29, 2006 ### Perturbation The velocity operator is usually defined by the following [tex]\vec{\mathbf{V}}=i[H,\mathbf{\vec{Q}}]$$

Where H is the Hamiltonian and Q the position operator.

4. Apr 29, 2006

### nrqed

However, this raises an interesting question (which might be what the OP had in mind but maybe not). All is fine when we just look at the maths, but let's say someone goes out in the lab and actually measure the momentum. At first sight, the only way to do this is to measure the positions at two times very close and that wreaks havoc with comapring with theory because the first measurement of position will collapse the wavefunction so it seems that whatever one gets by the measurement of the momentum can be compared in any meaningful way with what the theory predicts.

Maybe the only way out is to use a collision to measure the momentum instead of a displacement over a time interval times a mass.

Any thoughts?

Pat

5. Apr 29, 2006

### Staff: Mentor

Yes, I think at that scale the only practical way to measure momentum would be by way of scattering experiments. In fact, I think scattering experiments have been used to study the momentum distribution of atomic electrons, but atomic physics isn't my field so I don't have any specific references. I know that sort of thing is done in high-energy particle physics, for momentum distributions of quarks inside nucleons.

6. Apr 30, 2006

### pivoxa15

http://quantummechanics.ucsd.edu/ph130a/130_notes/node498.html
according to this website thre should be a factor of (1/hbar) in front of the operator.

What happens if your wavefunction has no complex numbers? You will get a velocity with a complex number. What would that mean?

Or since the Hamiltanian is equivalent to the energy operator (which have a factor of i so the two i cancels)should I use that instead of the Hamiltonian? But my wavefunction is independent of t so the average velocity will be 0. Just like the average momentum.

7. Apr 30, 2006

### pivoxa15

Yes, I am talking about the infinite 1D well. Thanks for correcting my error. I was getting expectation value confused with the sharp observable.

So as soon as I make a measurement of an electron, the wavefunction associated with that electron has dissapeared and a new one will emerge which means that the electron after the measurement is completely unrelated to the one before the measurement, even though there was always just one electron in the well. Is it like I have wiped out the 'electron's memory' by disturbing it? So it will start it fresh or like being born again after my measurement?

8. Apr 30, 2006

### pivoxa15

That is what I had in mind. So the only legimate way to measure momentum is from collisions hence p=srt(2mE). It means that it is (intrinsically) not possible to measure an electron's velocity. If so than how can we speak about the velocity of the electron if we can't measure it directly? It is accepted that the electron has no trajectory and for me an object can only have a velocity if they have a trajectory in space and time. So there is even more reason for disallowing the notion of velocity for an electron. Yet there is one and even an operator for it. How come?

9. Apr 30, 2006

### ZapperZ

Staff Emeritus
But think about it. What you said could be applied to ALL observables. And what exactly is a "direct" measurement? Everything, technically, is a measurement via detection/interaction.

In angle-resolved photoemission spectroscopy, the "simultaneous" measurement of electronic energy and momentum is made on a CCD detector (example: Scianta SES analyzer). This gives you a very good band structure dispersion of what the electron was doing (it's energy AND momentum parallel to the surface) right before it was emitted. If you do this on molybdenum, for example, you could even match the theoretical predictions, right down to seeing the phonon effects and electron-electron effects.

Now with that in mind, it is hard to convince me that one cannot measure such properties, and measure it with that degree of accuracy. If such values can't be extracted from experiment, then you will have a nasty time trying to explain what exactly are being measured in these experiments that agreed so well with theoretical descriptions.

Zz.

10. Apr 30, 2006

### Perturbation

There should be, yeah, from the Hamiltonian. I've just missed it out because I'm used to using units where h-bar=1.

The velocity operator is hermitian. It is the eigenvalues of these hermitian operators that correspond to physical observables. The eigenvalues of an Hermitian operator are always real.

11. Apr 30, 2006

### pivoxa15

I agree that
$$[H,\mathbf{\vec{Q}}]$$

will be real especially if my wavefunction is real. But I then have to multiply i to it which makes v imaginary.

But in the case of the infinite 1D well, the expected v just like expected p will be zero because of equal magnitude of negative and positive direction.

Last edited: Apr 30, 2006
12. Apr 30, 2006

### pivoxa15

So nowdays we can measure the energy and momentum of an electron to very high accuracy but does that automatically imply we can measure its velocity? If so than it means velocity is defined to be a consequence of energy and momentum. Is it?

In classical mechanics, velocity can always be associated with a trajectory of the particle. But in qm there is no such thing. So even though we can speak about the energy and momentum of a particle whether classically or quantum mechanically, we cannot do that with velocity. There is a difference between the velocity in classical mechanics and velocity in qm, namely a path. The former is intuitive and the latter unintuitive. If makes me wonder whether we should use the word velocity to describe atomic things because the use of the word 'automatically' make most people think about a path for that particle which is very misleading.

13. Apr 30, 2006

### masudr

Not really, no. In the Hamiltonian formulation, the momentum and position together describe the motion of the particle in the simplest way. And this translates to QM since Hamiltonian mechanics was quantized.

Also note that momentum does not always equal $m\vec{v}$. For example, a charged particle in an external EM field.

14. Apr 30, 2006

### ZapperZ

Staff Emeritus
Again, I have no idea if you're talking about the general principle, or something in particular. I can counter what you said again using examples from band structure of material. There is nothing, in principle, that prevents me from arriving at the group velocity of the electrons at a particular energy and momentum. All I need to know is find the effective mass of the electron using the dispersion relation, and I'm done.

Refer to P.D. Johnson et al. Phys. Rev. Lett. 87, 177007 (2001). You'll see a discussion on the different velocity of the band electrons for various doping level of a high-Tc superconductors.

Zz.

15. Apr 30, 2006

### Staff: Mentor

It makes the operator $\hat v$ imaginary. But its eigenvalues and expectation values, which are what we actually measure, are real. Note the momentum operator is also imaginary, in the position representation at least.

16. May 1, 2006

### pivoxa15

I get it now. I made the mistake of thinking it gave the actual velocity - where really it is just the velocity operator. That operator operates on the wavefunction to get an expected value of the obsevable. That expected value is always real because the domain we are integrating over is real. But velocity is not a sharp observable because momentum is not a sharp observable. So why did you mention eigenvalues in your post?

17. May 1, 2006

### pivoxa15

I am more concerned with the definition of velocity. I assume this definition should be the same whether macroscopically or microscopically. It seems that for you, the velocity does not have to be measured the traditional way of delx/delt but anyway that is consistent with the equations such as p/m=srt(2Em)/m.

Another thing I just realised is that why do we say there is no path for an electron? It could be that there is a path but we can never know it because of HUP. Therefore, an electron does have a definite velocity at any time and place but we can never know - we can only say its expected velocity and the standard deviation of it. Although when it is energetic enough such as when travelling in a TV tube, the standard deviation is much smaller around the mean and the HUP becomes less relavant.

18. May 1, 2006

### ZapperZ

Staff Emeritus
That isn't what I'm saying at all. All I did was to say that what you are trying to do does not work all the time. That is why I asked if you are doing this as a general principle or for something in particular. If you are doing this as a general principle, then I've just given you a situation where your general principle fail. If you are doing this for a particular situation (such as a free particle), then say so.

Notice why got into this thread in the first place. I was giving you an illustration where we DO have the ability to "directly" measure momentum of particles such as electrons (and holes in material). In fact, the raw data to do that is staring at you right now - it is in my avatar. The horizontal axis or cut is the momentum distribution of the outgoing photoelectrons.

... and btw, the velocity of an electron inside matter cannot be related as easily as your free-particle formulation. You are forgetting that E and p do not commute except for V=constant or zero.

Zz.

19. May 1, 2006

### Perturbation

Observable physical quantities correspond to the eigenvalues of operators. Expectation values are a statistical measure of the event we most expect (it's the same as the mean but for a random variable).

Just because there is an i in front of the commutator does not mean the velocity operator is not Hermitian. Whether something is Hermitian or not depends if it fulfils the following definition

$$\int \psi^* V\phi d^3x=\left[\int \phi^* V\psi d^3x\right]^*$$

20. May 1, 2006

### masudr

Yes. I personally like to write it as

$$\left\langle\phi|V|\psi\right\rangle = \left\langle\psi|V|\phi\right\rangle^*\Rightarrow\mbox{V is Hermitian}$$

since other definitions are restricted to specific kinds of Hilbert space.