Hamiltonian Math Problem: Finding the Time-Derivative of <p^2/2m>

In summary: I have tried it your way as well and gotten V(x) p^2/2m - p^2/2m V(x) for the commutator..You have solved this problem correctly.
  • #1
land
31
0
One more question. Sorry!

Here's the problem:

An electron moves in a straight line under the influence of a conservative force so that the Hamiltonian is [tex]H = \frac{p\wedge^2}{2m} + V(x)[/tex], where [tex]p\wedge[/tex] means the momentum operator and I think V(x) is the potential energy. I need to find an expression for [tex]\frac{d}{dt} <\frac{p^2}{2m}>[/tex].

Sigh. Anyone have any idea how to do this? I wish I could show you something that I've done to get started but I really don't have a clue. I do have an expression for the time-derivative of an operator.. but plugging this into it involves time derivatives of integrals over x and figuring out the commutator of H and p^2 / 2m, which involves doing calculations with the potential energy, and I have no idea how to deal with that. Thanks.
 
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  • #2
I don't really know the meaning of "I need to find an expression for..."
If you want to calculate something you need to know the wave function, they probably mean that you should do it to an arbitrary one (?)

Note that you have to calculate the time derivative of the kinetic energy expectation value. Calculate the total energy (you have to know the wave function for that), calculate the <V(x)> and the difference is the desired k.e exp. val.
I'm not sure though, sorry :blushing:
 
  • #3
The wave function isn't given, so I'm not sure what to do :( oh well, thanks!
 
  • #4
This is a nice question that helps to show the relationship between classical mechanics and quantum mechanics.

Have you covered the expression for the time derivative of expectation value of an operator?

As you say, you have to calculate [H, p^2/2m], but this isn't so bad.
 
  • #5
OK, I think I know what you're talking about. I did the commutator you were talking about and got [H, p^2/2m] = [tex]\frac{\hbar^2}{2m}\frac{\delta^2H}{\delta x^2}[/tex]

sound right?

So now I need to do the second derivative with respect to x of H.. and I'm not sure how to go about that, since I'm going to be taking the second derivative of a second derivative?
 
  • #6
k, I'm really stuck on this one. based on what I'm trying to prove (i realize i left this off, but i thought it would be elementary once i got the solution - stupid, i know), which is that in the classical limit [tex]\hbar \rightarrow 0[/tex], the expression reduces to [tex]\triangle K = \int Fdx[/tex], the work-energy theorem, I gather that my expression for d/dt <p^2/2m> must reduce to a double time integral integral of dV(x)/dx, which corresponds to the integral of force... but i have no idea how to get there. anybody have any hints or suggestions? :)
 
  • #7
Changed title of post to match what is going on here, because the original one wasn't very accurate.

Update before I go to bed: as I said in the last post, and as alluded to by George, I'm trying to show the relationship between the result and the one given by classical mechanics. This is where I'm stuck now:

I have [tex]\frac{d}{dt} <\frac{p^2}{2m}> = \frac{\hbar i}{2m}<\frac{\delta^2H}{\delta x^2}>[/tex]

This is after the commutator and setting the time derivative of p^2/2m to 0.

I have in my notes that you can set dH/dx equal to dV/dx.. but I still don't see how this is going to lead to the work-energy theorem. I'm also confused because my hbar is being multiplied by my expectation value.. so if I let it go to 0, the whole equaltion should go to zero, which doesn't make much sense at all. I could really use some help here:)
 
  • #8
land said:
I did the commutator you were talking about and got [H, p^2/2m] = [tex]\frac{\hbar^2}{2m}\frac{\delta^2H}{\delta x^2}[/tex]

This isn't the expression that I get.

[H , P^2] = [P^2/2m + V(x) , P^2] = ?
 
  • #9
Ah, in class we did it the way I was describing, just leaving H in there without explicity expressing it. I have tried it your way as well and gotten V(x) p^2/2m - p^2/2m V(x) for the commutator.. I still fail to see how that's going to lead to the answer, though, since the hbar is still outside... Unless I'm still doing something wrong, because my end result [tex]\frac{d}{dt} <\frac{p^2}{2m}> = \frac{\hbar i}{2m}<\frac{\delta^2V(x)}{\delta x^2}>[/tex] is still the same.
 
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  • #10
land said:
I have tried it your way as well and gotten V(x) p^2/2m - p^2/2m V(x) for the commutator

You need to do a little more work on the second term.
 
  • #11
Well, if you meant doing the product rule, I've tried that now and gotten a slightly different result (I messed up the sign before):

[tex][H,\frac{p^2}{2m}] = \frac{\hbar^2}{2m} ( 2V(x) \frac{\delta^2}{\delta x^2} + \frac{\delta^2 V(x)}{\delta x^2})[/tex]

but I fail to see how this is any more useful :(
 
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  • #12
land said:
Well, if you meant doing the product rule

Yes.

[tex][H,\frac{p^2}{2m}] = \frac{\hbar^2}{2m} ( 2V(x) \frac{\delta^2}{\delta x^2} + \frac{\delta^2 V(x)}{\delta x^2})[/tex]

Now the first term isn' right.

but I fail to see how this is any more useful :(

[tex]\triangle K = \int Fdx[/tex]

Can you change the integration variable to time?
 
  • #13
George Jones said:
Now the first term isn' right.

It isn't? I've done it three times now and gotten the same result each time..

And as far as changing the integration variable to time.. you mean, by Fourier transform? I'm lost.
 
  • #14
land said:
It isn't? I've done it three times now and gotten the same result each time..

Without seeing your work, I can't tell where you went wrong.

And as far as changing the integration variable to time.. you mean, by Fourier transform? I'm lost.

No, I just mean write F(x) as F(x(t)), where x(t) is the path of the particle, and write dx = (dx/dt) dt.

Then, what is

[tex]\frac{d}{dt} \left( \Delta K \right)?[/tex]
 
  • #15
ok, here's what I did:

[tex][\frac{p^2}{2m} + V(x)](\frac{p^2}{2m}) - \frac{p^2}{2m}[\frac{p^2}{2m}+V(x)][/tex]
.
.
[tex]\frac{p^4}{4m^2} + V(x)\frac{p^2}{2m} - \frac{p^4}{4m^2} - \frac{p^2}{2m}V(x)[/tex]
.
.
[tex]\frac{\hbar^2}{2m}V(x)\frac{\delta^2}{\delta x^2} + \frac{\hbar^2}{2m}[V(x)\frac{\delta^2}{\delta x^2} +\frac{\delta^2 V(x)}{\delta x^2}][/tex]
.
.
[tex]\frac{\hbar^2}{2m}[2V(x)\frac{\delta^2}{\delta x^2} + \frac{\delta^2 V(x)}{\delta x^2}][/tex]and... would [tex]\frac{d}{dt} \left( \Delta K \right)[/tex] just be F(x(t))?
 
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  • #16
Good!

land said:
[tex]\frac{p^2}{2m}V(x)[/tex]

When expanding this term, a little care is needed. Do it one step at a time, and, if it helps, put in a [itex]\psi[/itex].

[tex]\frac{p^2}{2m}V\psi = \frac{p}{2m} \left{ p \left( V \psi \right) \right}[/tex]

Also, I think there is a sign mistake in the first term of the third line.
 
  • #17
Ahhhh, I see. Now I'm getting [tex]\frac{\hbar^2}{2m}[\frac{\delta^2 V}{\delta x^2} + 2\frac{\delta V}{\delta x}\frac{\delta}{\delta x}][/tex]

Still not seeing it :(
 
  • #18
land said:
Ahhhh, I see. Now I'm getting [tex]\frac{\hbar^2}{2m}[\frac{\delta^2 V}{\delta x^2} + 2\frac{\delta V}{\delta x}\frac{\delta}{\delta x}][/tex]

What is the time derivative of the expectation value of the first term?

Write the last term in terms of p.
 
  • #19
You mean actually go through and calculate [H, d^2V/dx^2]? Or is there something I'm supposed to be seeing here?

Write in terms of p? I'm not sure what you mean, unless it's replace d/dx with ip/hbar.
 
  • #20
land said:
You mean actually go through and calculate [H, d^2V/dx^2]? Or is there something I'm supposed to be seeing here?

No, it's much simpler. The expectation value is an an integral, and since V doesn't depend on time, the only time dependence is in the [itex]\Psi[/itex] parts. But since V'' is a scalar it can be move throught a [itex]\Psi[/itex], giving [itex]V'' \Psi \Psi *[/itex] under the integral.

What is [itex] \Psi \Psi *[/itex]? Does the integral have any time dependence?

Write in terms of p? I'm not sure what you mean, unless it's replace d/dx with ip/hbar.

Yes.
 
  • #21
George Jones said:
No, it's much simpler. The expectation value is an an integral, and since V doesn't depend on time, the only time dependence is in the [itex]\Psi[/itex] parts. But since V'' is a scalar it can be move throught a [itex]\Psi[/itex], giving [itex]V'' \Psi \Psi *[/itex] under the integral.

What is [itex] \Psi \Psi *[/itex]? Does the integral have any time dependence?

Ah. so the integral is just 1, right? So the expectation value is just V''.. and the time derivative is 0?

but why just take the time derivative of the expectation value? i know it wasn't arbitrary..
 
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  • #22
land said:
Ah. so the integral is just 1, right? So the expectation value is just V''.. and the time derivative is 0?

Right. (extra characters)
 
  • #23
but what does that have to do with anything? you can just disregard that part then? why?
 
  • #24
but why just take the time derivative of the expectation value? i know it wasn't arbitrary..

Edit: Oops, i made a mistake.

would [tex]\frac{d}{dt} \left( \Delta K \right)[/tex] just be F(x(t)
it

What happened to the dx/dt?

Also, write F in terms of the potential.
 
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  • #25
Ah.. the F = -dV/dt. and the dx/dt is still there I suppose..
 
  • #26
Forget about time derivative of the expectation value of V''. Just do the integral, and assume that the force on the particle dies at infinity.
 
  • #27
What integral, [itex]V'' \Psi \Psi *[/itex] from -infinity to infinity?
 
  • #28
land said:
What integral, [itex]V'' \Psi \Psi *[/itex] from -infinity to infinity?

Sorry, I had to go.

I was trying to force something to zero that isn't zero.

Go back to

[tex]\frac{\hbar^2}{2m} \left[ \frac{d^2 V}{dx^2} + 2\frac{d V}{d x}\frac{d}{d x} \right].[/tex]

This is a little too compact; went a little too far with the product rule.

[tex]\frac{\hbar^2}{2m} \left[ \frac{d^2 V}{dx^2} + 2\frac{d V}{d x}\frac{d}{d x} \right] = \frac{\hbar^2}{2m} \left[\frac{d^2 V}{dx^2} + \frac{d V}{d x}\frac{d}{d x} + \frac{d V}{d x}\frac{d}{d x}\right][/tex]

Now, use the product rule to write the first two terms as d/dx of something.

Ah.. the F = -dV/dt. and the dx/dt is still there I suppose..

Not quite, F = -dV/dx
 
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1. What is a Hamiltonian Math Problem?

A Hamiltonian Math Problem is a type of mathematical problem that involves calculating the time-derivative of a specific formula or equation, known as the Hamiltonian. This problem is commonly encountered in the field of physics and is used to describe the dynamics and motion of a system.

2. What is the formula for the Hamiltonian?

The Hamiltonian is typically denoted by the letter H and is defined as the sum of the kinetic energy (T) and potential energy (V) of a system: H = T + V. In the case of the specific problem mentioned, the Hamiltonian is given by the formula p^2/2m, where p represents the momentum of the system and m is the mass.

3. How is the time-derivative of the Hamiltonian calculated?

The time-derivative of the Hamiltonian is calculated using the Hamilton's equations of motion. These equations involve taking the partial derivatives of the Hamiltonian with respect to the variables in the system, such as position and momentum. Once these equations are solved, the time-derivative of the Hamiltonian can be obtained.

4. What is the significance of solving the Hamiltonian Math Problem?

Solving the Hamiltonian Math Problem allows us to understand the behavior and dynamics of a system in terms of its energy. This can provide valuable insights and predictions about the system's motion and interactions. It is also an important tool in many fields, such as classical mechanics, quantum mechanics, and statistical mechanics.

5. Are there any real-world applications of the Hamiltonian Math Problem?

Yes, the Hamiltonian Math Problem has many real-world applications, particularly in the fields of physics and engineering. It is commonly used in the study of systems such as atoms, molecules, and particles, and has applications in fields such as astrophysics, fluid dynamics, and quantum computing. It is also used in the development of mathematical models for complex systems.

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