# Momentum Conserved?

1. Feb 3, 2004

### Crashdowngurl

Is it possible for momentum to be conserved in one direction and not in other directions? Explain your answers with reasons.

This was the question given to my classmates and I to solve. However my group and I had several very different takes on it, and I was wondering if someone could justify that I was right or explain perhaps why I am not right.

My take on it

Quantum mechanics proves that momentum is conserved. The assumption is that space is homogeneous in all directions - this leads to conservation of momentum. If I were to take everything in the universe, and move it to the right, then the laws of physics should still be the same. That means momentum is conserved in the x-direction. Similarly for whatever direction you want to show conservation of momentum, you would displace the universe in that direction and argue from homogenity of space that the physics stays the same. What happens is because the physics is symmetrical under the operation of moving it over a little (the Hamiltonian commutes with the displacement operator), and hence the momentum operator, which is the derivative of the displacement operator with respect to position, is also commutable with the Hamiltonian. So your state will always return the same eigenvalue (the value of definite momentum in your case) at anytime:

P[H[State 1]]=H[P[State 1]]=H[momentum*State 1]=
momentum*H[State 1]=momentum*State 2

(there's one detail, since the Hamiltonian H is the derivative of the time operator, and not the time operator itself, but you can show that if H commutes with another operator than U - the time operator - commutes with the operator too]

Similarly conservation of angular momentum is a consequence of the isotropy of space (if I rotate the whole universe then the physics is the same), and conservation of parity is ambidextrousnes of space (if I make the mirror world of the universe - parity is actually no conserved for beta decay), and conservation of energy is...well I'm not sure I can make up a catchy label for this one - how about I call it the "patience" of space (if move time backwards a little, then the laws of physics are still the same).

Classmates take on it

Law of conservation of momentum: “The momentum of any closed, isolated, system does not change”

It is possible for momentum to be conserved in one direction and not in the other. For example, If I were in an “isolated system” roller-blading, with no help of external forces, and were to slide into someone else in that system who was at rest, we would continue with the same momentum. If I were going 4.4 m/s, the momentum would be split between the two bodies in the system, but the momentum would remain the same as it was before I pushed the other body.

In order for the momentum to be conserved in that one direction however, no external forces could have been present. If an external force acts, the system would change, and the momentum would not be conserved.

2. Feb 10, 2004

### Tom Mattson

Staff Emeritus
First, this is not about quantum mechanics. The law of conservation of momentum holds in classical mechanics, too.

That is basically right. But usually we do not work with the universe in total, but rather with physical systems, and we do not assume that all physical systems are spatially homogeneous. Rather, we say that for physical systems that are spatially homogeneous, the laws of physics are unchanged under spatial translations.

This discussion of commutators in quantum mechanics is not necessary. Also, it is not quite correct. When you say that "the Hamiltonian commutes with the displacement operator...and hence the momentum operator...is also commutable with the Hamiltonian", the latter does not follow from the former. In fact, I don't see how any Hamiltonian could possibly commute with the displacement operator, because the displacement operator does not commute with p2/2m, which is the kinetic part of the quantum mechanical Hamiltonian.

Can anyone here think of an example of a Hamiltonian for which [H,x]=0?

Minor quibble: There is no "time operator" T-hat. You mean the "time evolution operator" U(t,t0).

Major quibble: H is not the derivative of U(t,t0). H is the time derivative of the wavefunction, modulo a factor of i*(hbar).

OK

In the currently accepted terms, we say that the energy of a system is conserved if its Lagrangian is invariant under time translations.

That is correct. By definition, F=dp/dt, and so for any direction in which F has a nonzero component, p will have a nonzero time derivative and thus will change.

And p will not be conserved in only those directions in which F has a nonzero component.

The basic difference between your approach and that of your classmates is that they are appealing to the definition of momentum, while you are appealing to Noether's theorem. Also, your answer does not take into account the fact that momentum is a vector, which is really at the heart of the issue.

The only other suggestion I would make would be for you to omit the discussion of quantum theory, as it is not necessary.

edit: fixed quote bracket

3. Feb 10, 2004

### RedX

But all systems are spatially homogeneous when nothing external is happening or if the external fields are constant or if the system is not affected by the external field.

Let U be the time evolution operator, H be the Hamiltonian, and O a generic operator, and W be the state:

U=1-(i/h)H(dt)

OUW=UOW

O(1-(i/h)H(dt))W=(1-(i/h)H(dt))OW

OHW=HOW
OH=HO

So if an operator commutes with the time evolution operator, it commutes with the Hamiltonian.

I think it's the derivative. You've got:

U=1-(i/h)H(dt)

It's the derivative of U(t,t0) at t=t0. It's a Taylor expansion.

F=dp/dt is not the definition of momentum. They're appealing to Newton's 2nd law. Even if you use QM to show that -dV/dx=d<p>/dt, you're still not defining momentum. I think you have to define momentum as the eigenvalue you get from the momentum operator or that <p>=m*d<x>/dt.

I think she wanted to prove that momentum is conserved, and if you prove that it's conserved in any direction, you prove that it's conserved in every direction.

4. Feb 11, 2004

### Tom Mattson

Staff Emeritus
The point is that some systems are not spatially homogeneous, and that we cannot use "homogeneity of space" to determine the conserved quantities of those systems. We have to rely on those transformations that leave the Lagrangian invariant, which demands consideration of the system in particular, rather than space in general.

Yes, I know all that. But she didn't say that! She said that if the Hamiltonian commutes with the displacement operator, then the momentum operator also commutes with the Hamiltonian.

In other words, you proved that [U,O]=0 implies that [H,O]=0, which I agree with. She said that [H,x]=0 implies that [H,p]=0.

I know it is the Taylor expansion. However, (and as you youself noted), the 1st order coefficient of the Taylor expansion is proportional to the derivative evaluated at the center of the series. That is not the same as "the derivative". That's like saying that the derivative of eax is equal to a.

You're right, it's the definition of force. My mistake.

Not for all physical systems it isn't.

edit: typo

Last edited: Feb 11, 2004
5. Feb 11, 2004

### krab

No. We are sitting here on earth and we have a gravitational force acting vertically but not horizontally. That means when we do experiments, momentum is conserved in the horizontal directions but not in the vertical direction.

6. Feb 11, 2004

### RedX

Say you have a generic system, and the only thing external going on is a constant gravitational field in the z-direction. As Krab said, momentum is not conserved in the z-direction - and that makes a lot of sense. However, it shouldn't matter if I displace the entire system in any given direction, since the field is constant everywhere. That should imply momentum is conserved - I could provide the proof that Feynman gives in his lectures, but I'm guessing you guys already know it. So I guess common sense is telling me that momentum isn't conserved, but quantum mechanics is telling me it ought to be.

I too thought that commutivity of the Hamiltonian and displacement operator implied commutivity of the Hamiltonian and momentum operator.

It goes something like:

Let D be the dispacement operator, P the momentum operator, H the Hamiltonian, and W the quantum state

D=1+(dx)(i/h)P

HDW=DHW
H(1+(dx)(i/h)P)W=(1+(dx)(i/h)P)HW
HPW=PHW
HP=PH

7. Feb 12, 2004

### krab

Yes, the field is constant but it is not the field that appears in H; rather it is the potential, and the potential is mgz, which clearly varies with z.

8. Feb 12, 2004

### Tom Mattson

Staff Emeritus
Krab already answered that one for me (see the post below yours).

A-HA! What you (and I presume Crashdowngurl) are calling "the displacement operator" is what I call "the translation operator". When I see the term "displacement operator", I equate that with "position operator".

OK, objection withdrawn.

9. Feb 15, 2004

### RedX

Feynman calls it the displacement operator. I guess translation operator is common terminology. I call position operator the position operator.