How to Prove d<L>/dt = <N> for a Particle in Potential V(r)?

[Tuneman]

The question reads: Prove that for a particle ina potential V(r) the rate of change of the expectation value of the orbital angular momentum L is equal to the expectation value of the torque:
d<L>/dt = <N>
where N = r x (-grad V)
Now I was thinking the key of this problem was messing with the Ehrenfest's theorem until it looked like the first equation. Obviously the r is going to come from substituting r x p for L. Any help would be appreciated, thanks.

I simplified it to this:
d/dt(L) = i/(hbar) ( p^2/2m + V , -i(hbar) x (grad) ) + ( dr/dt x p ) + (r x dp/dt)and I know dr/dt points in the same direction as p, so dr/dt x p is 0.So I simplified it down to this,
d/dt(L) = i/(hbar) ( p^2/2m + V , -i(hbar) x (grad) ) + (r x dp/dt)

and I need everything on the left except for the last term to equal zero.

 

[Galileo]

If you see the time derivative of the expectation value of an operator, you should be reminded of the QM 'equation of motion'. For time independent operators I think it was something like:

\frac{d}{dt}\langle Q \rangle =\frac{\hbar}{i}\langle [H,Q]\rangle

 

[Tuneman]

ya i know, I expanded that out where H = T + U, and Q equaled L, but what to do with that, I don't know.

Edit: Acutally the theorem I looked up also has a "+ (dL/dt)" <---partial deriviative. So I think those high school algebra steps i took at the beginning have nothing to do with solving the problem. I think I have to go back to square 1.

Also, there is a (b) part of the problem, that says show that d<L>/dt= 0 for any spherically symmetric potential. I haven't even looked at that part yet, not sure exactly how to solve that either..

 

[Tuneman]

any help would really be appreciated, I put a lot of time into this problem, and haven't gotten very far.

 

[Physics Monkey]

A useful operator identity is [AB,C] = A [B,C] + [A,C] B. Try applying this formula to the first component of the angular momentum operator, you will find that the resulting commutators are easy to evaluate. You can then obtain similar expressions for the other components of angular momentum by cyclically permuting indices.

 

[Tuneman]

when you say "the first component" of the angular momentum operator, are you talking about the operator in the equation I mentioned, or the sperate operators for each dimension of angular momentum?

 

[Physics Monkey]

The angular momentum operator is a vector, so when I say the first component I mean the first component of that vector i.e. L_x = y p_z - z p_y.

 

[Tuneman]

perhaps this seems like a stupid question but wouldn't I need to have 2 of the angular momentum components in order to apply that formulaa? For example:

[Lx,Ly] = ihLz

 

[Physics Monkey]

You want to calculate the time derivative of the expectation value of angular momentum. Galileo gave you a formula and so you need to calculate [L_x,H]. What's the problem?

 

[Tuneman]

hmm interesting, and so i am guessing h/i = A, H = B , Lx = C if I were to apply it to your formula

Edit: no, nevermind h/i is a cosntant, so you're saying have [ y p_z - z p_y , H ] ?

sry if these questions sound stupid, but if you knew how little our teacher taught us, and how poor this book is at explaing commuators...

 

[Physics Monkey]

Yes, you've got it. Now first use the simple identity [A+B,C] = [A,C] + [B,C] to separate the commutator into two terms, then you can apply the identity I told you above to each term. This is useful because the commutator of position or momentum with the Hamilitonian is easier to evaluate (I suspect you already know the results in fact).

 

[Tuneman]

Yes I think I do, I know
[ H , y ] = Hy-yH
and
[ H , P_z ] = HP_z - P_zH
are those the ones you mean (along with the other permutation of the idecies)?
Edit: I did the algebra/whatever and i got this
(Hbar)/ i { [ ( Hy-yH ) P_z + y ( HP_z - P_zH ) ] - [ ( Hz - zH ) P_y + z ( HP_y - P_yH ) ] }

 

[Galileo]

Correct, although written a bit more neatly as (omitting the constant factor hbar/i):

[H,y]p_z+y[H,p_z]-[H,z]p_y-z[H,p_y]
Let's look at, say, the first commutator [H,y]. How would you evaluate this?

You know that H=p^2/2m+V(r) (are you given that V depends only on the radial distance r=\sqrt{x^2+y^2+z^2}?)
With the commutation relations [x_i,p_j]=i\hbar\delta_{ij} you can evaluate most terms.
Something like [V,p_z] may require a little additional calculation.

 

[Tuneman]

Ok so Ill expand the hamiltonian out and use [A+B,C] = [A,C] + [B,C], where A = p^2/2m , B = V(r) . One question, what do I do about the momentum vector p^2 do I have to break that up into its components?

 

[Tuneman]

sry if these questions seem simple, however if you knew how little the professor teaches us, you'd understand.

 

[Galileo]

Do I have to break that up into its components?

You must do whatever you must do to solve the problem. If you think expanding out the commutator helps you to evaluate the commutator, do so!
There doesn't seem to be much choice in this case, although you can speeden up the calculation by noting that [p^2,p_i]=0, [V,x_i]=0. I hope these identities seem obvious.

 

[Tuneman]

ya I had tried that, and I had used those identies last night, however, I'm still getting stuck, I have no idea how I am going to get from this combination of commutators to the cross product d&lt;L&gt;/dt = r x (- /nabla V) where x is a cross product. Should the answer be obvious to me?

Here is where I am currently at :

P_z/2m[p^2_z, y] + P_z/2m [P^2_x,y] - P_y/2m[P^2_y, z] -P_y/2m[P^2_x, z] + y [V,P_z] - z[V, P_y]

 

[Galileo]

What we're doing is equating the components of the left and right side. So instead of using the vector equation with the cross-product we treat it as a set of 3 equations. One for each component. (There may be other ways to do it, but this is one is relatively easy and straightforward).

You've got a lot of terms in the expression. Many of them are zero, like [p_x,y]. Do you see why?

Also, you should be able to evaluate [V,P_z]. Let it act on a test function and see what happens.

You're close to the answer, although it may not look like it

BTW, there was a slight error in the motion equation I posted. The factor \hbar/i should've been i/\hbar, so it's

\frac{d}{dt}\langle Q \rangle =\frac{i}{\hbar}\langle [H,Q]\rangle

 

[Tuneman]

I am not sure why those terms, are zero, I was under the impression the only terms that could equal zero were the ones that were like [P_x, x] or [P_y, y] which I did set to zero in the equation in my previous post.

 

[Galileo]

Nonono. These are the ones that are not zero! They do however cancel out in the expression, so that's a coincidence.

Surely you've seen before that [x,p_x]=i\hbar ??

You can demonstrate it by taking an arbitrary function f to have [x,p_x] act upon it:

[x,p_x]f=x\frac{\hbar}{i}\frac{\partial}{\partial x}f-\frac{\hbar}{i}\frac{\partial}{\partial x}(xf)=\frac{\hbar}{i}\left(x\frac{\partial}{\partial x}f-f-x\frac{\partial}{\partial x}f\right)=(i\hbar)f
So [x,p_x]=i\hbar. All I've used is the product rule.

You can work out all terms this way, but I figured you've seen this before. In condensed notation the canonical ones are (see post 13):

[x_i,x_j]=[p_i,p_j]=0

[x_i,p_j]=i\hbar \delta_{ij}

where i,j=x,y,z.

 

[Tuneman]

Well here's the thing, I have:

p_z/2m([p^2_x,y] + [p^2_y,y] + [p^2_z, y]) + y[V,p_z] - p_y/2m([P^2_x,z] + [p^2_y, z] + [p^2_z,z]) - z[V,P_y] \\

so the therefore there is a

[p^2_x, y] = i\hbar \delta_{xy}
[p^2_z, y] = i\hbar \delta_{zy}
[p^2_x, z] = i\hbar \delta_{xz}
[p^2_y, z] = i\hbar \delta_{yz}

along with:
P_z/2m[P^2_y, y]
and
-P_y/2m[P^2_z, z]

I am not sure how those will cancel out since the P_z, and P_y are different
so the terms are all different, not to mention that I am multiplying the first two by P_z/2m, and the last two by P_y/2

 

[Galileo]

You do know that \delta_{ij} is the kronecker delta, right? It's equal to zero if i\not=j and 1 otherwise.

Also, consider [p_x,y] this way. Does it matter whether you differentiate wrt to x and then multiply by y, or whether you multiply by y and THEN differentiate with respect to x?

 

[Tuneman]

sry double post, see the below one.

 

[Tuneman]

Ah I think I see where this is going if I use the the relations:

[x,p_x]=i\hbar

and the kronecker delta relations to my equation:

p_z/2m([p^2_x,y] + [p^2_y,y] + [p^2_z, y]) + y[V,p_z] - p_y/2m([P^2_x,z] + [p^2_y, z] + [p^2_z,z]) - z[V,P_y] \\

I come out with

(P_z/2m)i\hbar - (P_y/2m)i\hbar + y[V,P_z]-z[V,P_y]

Which looks a awful lot like L_x except for the last two terms, and the coefficents of the last two terms should be on the first two terms. This should be able to lead me somewhere

 

[Galileo]

I think you lost a factor somewhere. Notice that all that survives is:

\frac{1}{2m}(p_z[p_y^2,y]-p_y[p_z^2,z])+y[V,p_z]-z[V,p_y]

Now use [AB,C]=A[B,C]+[A,C]B to evaluate the [p_y^2,y] like terms.

After that, use the test function method to find out what [V,p_y] is, so you can find the last 2 terms.

 

[Tuneman]

Ok so I got the first two terms to cancel each other out. Then applying the test function, I did this:

y[V,p_z]=[V,p_z]y=V\frac{\hbar}{i}\frac{dy}{dz} - y \frac{\hbar}{i}\frac{dV}{dz} = 0 - y \frac{\hbar}{i}\frac{dV}{dz} does that seem right?

 

[Tuneman]

Ok so I got the first two terms to cancel each other out. Then applying the test function, I did this:

y[V,p_z]=[V,p_z]y=V\frac{\hbar}{i}\frac{dy}{dz} - y \frac{\hbar}{i}\frac{dV}{dz} = 0 - y \frac{\hbar}{i}\frac{dV}{dz} did I do that right because then my entire thing becomes,

- y \frac{\hbar}{i}\frac{dV}{dz} + z \frac{\hbar}{i}\frac{dV}{dy}

then adding in the \frac{i}{\hbar} that was in front of this entire thing the whole time, I cancel those hbars and i's out and get

\frac{dL_x}{dt} = z\frac{dV}{dy} - y\frac{dV}{dz}

 

[harsh]

I might be wrong about this, but ist d<L>/dt = 0, if the potential is spherically symmetric? Its proportional to the expectation value of the commutator of H and L. That is, <[H,L]>, which is zero?

- harsh

 

[Tuneman]

thats actually part (b) of the problem. Its to prove that it equals zero.however I am having trouble equaiting my last equation to

\frac{dL_x}{dt} = r \X\ (- \nabla V)

 

[Galileo]

\frac{dL_x}{dt} = z\frac{dV}{dy} - y\frac{dV}{dz}

Remember that, in the Schordinger picture, it's the expectation value that is important, because the operators themselves don't change in time.

So you get:

\frac{d}{dt}\langle L_x \rangle = \langle -y\frac{\partial V}{\partial z}+z\frac{\partial V}{\partial y} \rangle

Does the term in brackets look like something familiar? Remember the goal is to show that it should equal the x-component of \langle \vec r \times (-\vec \nabla V)\rangle. Does it?

And V(\vec r) doesn't have to be radially symmetric. Although you'll find \vec r \times \vec \nabla V=0 if it does.

 

[Tuneman]

ya that is the x component, of r \ X\ (-\nabla V) and then I can simply argue if a add the three components, L_x, L_y, L_z knowing r = x,y,z, I can say \frac{dL}{dt} = r \X\ (- \nabla V)

Wow...that makes total sense... Ty very much for helping me solve the problem, I really appreciate it.

 

[Galileo]

Yep, since similar results hold for L_y and L_z. You can just permutate the indices cyclicly (is that a word?)

\vec L is just a operator with 3 components: \vec L=L_x \vec i+L_y\vec j +L_z \vec k, so you can just equate the components of both sides.

 

[Tuneman]

awesome, thanks for the help!

and to show that to show that for a sphyrically symmetric potential \frac{d&lt;L&gt;}{dt} = 0, someone mentioned to me to try this

\frac{d&lt;L&gt;}{dt} = [H,L] + \frac{dh}{dt}

where the right term on the right hand side equals zero becuase the hamiltonian isn't time dependent, and then I guess I am left off where i started in the previous problem. Should I be thinking of this equation, or should I be trying a different approach.

 

[Galileo]

The equation you derived is general, ofcourse.

\frac{d}{dt}\langle \vec L \rangle = \langle \vec r \times (-\vec \nabla V) \rangle

So try to evaluate the right hand side for the case where V=V(r), r=|\vec r|.

Hint: Use \vec \nabla in spherical coordinates.

 

[Tuneman]

Ah I see!
r \times -\nabla V = r(r \times r)(\frac{-dV}{dr})+ r(r\times \theta)\frac{1}{r}(\frac{-dV}{d\theta})+r(r\times\phi)\frac{1}{rsin\theta}(\frac{-dV}{d\theta})
Where the things getting cross multiplied are the unit vecotrs (i don't know how to put the hats on them) and of course those are partial deriviatives.
The first term has r x r, which is of course 0, and the other two terms have me differentiating V(r) with respect to the two angular coords, so of course they are 0! so the whole thing equals zero!

 

[Galileo]

Yep. The quick way of seeing this intuitively is by recalling that \vec \nabla V is the vector which, at eacht point, points in the direction of maximum increase of V. If V is spherically symmetric, what other direction can this be but radially outward? In the tangential direction it must be zero.
So \vec \nabla V(r) points in the direction of \vec r everywhere, so its cross product with \vec r is zero everywhere.

 

[harsh]

Actually, you could have used the equation is suggested. The commutator [H,L] is zero, and <dl/dt> is obviously zero. Its a nifty little formula, and its very useful.

- harsh