Ehrenfest Theorem: deriving d<p>/dt = <-dV/dx>

[buffordboy23]

I am having trouble with this derivation. This is where I am at:

\frac{d \left&lt; p \right&gt;}{dt} = \int \left[ \left( -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi^{*}}{\partial x^{2}} + V\Psi^{*} \right) \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi^{*}}{\partial x}\left( \frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi}{\partial x^{2}} - V\Psi \right) \right] dx + \int \Psi^{*}\left(-\frac{\partial V}{\partial x} \right) \Psi dx

\frac{d \left&lt; p \right&gt;}{dt} = \int \left( H\Psi^{*}\frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^{*}}{\partial x}H\Psi \right) dx + \int \Psi^{*}\left(-\frac{\partial V}{\partial x} \right) \Psi dx

I am guessing that this is correct so far; it looks like the terms under the first integral will cancel out nicely somehow and the second integral has the term of interest. If this is incorrect, then I will post my previous steps of the derivation.

Is the Hamiltonian commutative? At this point I am confused and do not know how to proceed.

EDIT: I just thought of something. Can I say that since H\Psi = E\Psi, where E represent the energy and is a constant. Therefore, I can pull this constant outside the integral and apply an integration by parts to get

\frac{d \left&lt; p \right&gt;}{dt} = E \int \left(\Psi^{*}\frac{\partial \Psi}{\partial x} - \frac{\partial \Psi^{*}}{\partial x}\Psi \right) dx + \left&lt;-\frac{\partial V}{\partial x} \right&gt; = \left E\Psi^{*}\Psi \right|^{+ \infty}_{- \infty}+ \left&lt;-\frac{\partial V}{\partial x} \right&gt;= \left&lt;-\frac{\partial V}{\partial x} \right&gt;

since \Psi goes to zero at infinity.

EDIT2: This shouldn't work either because of the minus sign between the terms in the first integral.

 

[Dick]

You DON'T want to use H(psi)=E*psi. The result is supposed to be true even for wavefunctions that aren't energy eigenfunctions. And H is not 'commutative', it's hermitian. And I think that minus sign is an error. Check it.

 

[buffordboy23]

Yes, you are right that the minus sign should be a plus sign.

<br /> \frac{d \left&lt; p \right&gt;}{dt} = \int \left[ \left( -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi^{*}}{\partial x^{2}} + V\Psi^{*} \right) \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi^{*}}{\partial x}\left( -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi}{\partial x^{2}} + V\Psi \right) \right] dx + \int \Psi^{*}\left(-\frac{\partial V}{\partial x} \right) \Psi dx<br />

So, if I shouldn't invoke H\Psi = E\Psi, then I guess using integration by parts with the product rule should lead to the cancellation of this first integral. Does this make sense? Also, when using integration by parts, I imagine that I should assume that V depends on "x".

 

[Dick]

Yes, you are right that the minus sign should be a plus sign.

<br /> \frac{d \left&lt; p \right&gt;}{dt} = \int \left[ \left( -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi^{*}}{\partial x^{2}} + V\Psi^{*} \right) \frac{\partial \Psi}{\partial x} + \frac{\partial \Psi^{*}}{\partial x}\left( -\frac{\hbar^{2}}{2m}\frac{\partial^{2} \Psi}{\partial x^{2}} + V\Psi \right) \right] dx + \int \Psi^{*}\left(-\frac{\partial V}{\partial x} \right) \Psi dx<br />

So, if I shouldn't invoke H\Psi = E\Psi, then I guess using integration by parts with the product rule should lead to the cancellation of this first integral. Does this make sense? Also, when using integration by parts, I imagine that I should assume that V depends on "x".

If you do an integration by parts on the first integral you'll get another copy of the second integral. So somehow that's in there twice. Maybe you should post your results leading to this. You should be able to show d<p>/dt=<[p,H]>/(i*hbar) and [p,H]=-dV/dx. And, yes, sure V depends on x.

 

[buffordboy23]

Like you said, I also saw how the copy of the last integral surfaces when using integration by parts.

From the beginning, we have:

\left&lt; p \right&gt; = -i\hbar \int \Psi^{*}\frac{\partial \Psi}{\partial x} dx

Taking the derivative with respect to time gives

\frac{d \left&lt; p \right&gt;}{dt} = -i\hbar \int \frac{\partial}{\partial t}\left( \Psi^{*}\frac{\partial \Psi}{\partial x}\right) dx= -i\hbar \int \left[ \frac{\partial \Psi^{*}}{\partial t}\frac{\partial \Psi}{\partial x}+\Psi^{*}\frac{\partial}{\partial t}\left(\frac{\partial \Psi}{\partial x}\right) \right] dx

Since \partial \Psi/\partial x = (ip/h)\Psi, substitution gives,

\frac{d \left&lt; p \right&gt;}{dt} = -i\hbar \int \left[ \frac{\partial \Psi^{*}}{\partial t}\frac{\partial \Psi}{\partial x}+\Psi^{*}\frac{\partial}{\partial t}\left(\frac{ip}{\hbar}\Psi\right) \right] dx

I guess this is where things for me start to get questionable. I made this substitution because I wanted to get \partial \Psi/\partial t to make substitutions from the Schrodinger equation; I assumed that the order in which the position and time derivatives is applied is important(?), so I had no other choice. I also believe that "p" is not commutative because it is the differential operator. For my previous post I assumed that "p" is differentiable with respect to time; I now think this is an error(?). Any thoughts are appreciated.

 

[nrqed]

Like you said, I also saw how the copy of the last integral surfaces when using integration by parts.

From the beginning, we have:

\left&lt; p \right&gt; = -i\hbar \int \Psi^{*}\frac{\partial \Psi}{\partial x} dx

Taking the derivative with respect to time gives

\frac{d \left&lt; p \right&gt;}{dt} = -i\hbar \int \frac{\partial}{\partial t}\left( \Psi^{*}\frac{\partial \Psi}{\partial x}\right) dx= -i\hbar \int \left[ \frac{\partial \Psi^{*}}{\partial t}\frac{\partial \Psi}{\partial x}+\Psi^{*}\frac{\partial}{\partial t}\left(\frac{\partial \Psi}{\partial x}\right) \right] dx

Since \partial \Psi/\partial x = (ip/h)\Psi, substitution gives,

You can't do that since Psi is not necessarily a momentum eigenstate.

All you have to do are integrations by parts (after using the time dependent schrodinger equation)

 

[buffordboy23]

I see. This makes sense. I complicated the problem. It seemed like a great substitution, but I never knew that I could not do such things in QM. However, the mathematics now gets in the way from further progress. I see two partial derivatives, position and time, acting on \Psi; can I rip the position derivative off although the time derivative is adjacent to its left? I want to crush this derivation..

 

[nrqed]

I see. This makes sense. I complicated the problem. It seemed like a great substitution, but I never knew that I could not do such things in QM. However, the mathematics now gets in the way from further progress. I see two partial derivatives, position and time, acting on \Psi; can I rip the position derivative off although the time derivative is adjacent to its left? I want to crush this derivation..

Use Schrodinger's equation to replace the time derivatives. After that there will be no time derivatives left, just terms with derivatives with respect to x and the potential.

 

[Dick]

I see. This makes sense. I complicated the problem. It seemed like a great substitution, but I never knew that I could not do such things in QM. However, the mathematics now gets in the way from further progress. I see two partial derivatives, position and time, acting on \Psi; can I rip the position derivative off although the time derivative is adjacent to its left? I want to crush this derivation..

The partial derivatives with respect to t and x do commute. Actually, you made me look this up. It's called Clairaut's theorem or Young's theorem or something. There are mathematically pathological cases in which they don't. But they aren't of interest in physics. Assume partial derivatives with respect to the independent variables commute.

 

[buffordboy23]

Thanks for the great feedback. I am learning a lot from this thread. Here is my completed derivation for review.


We know that

\left&lt; p \right&gt; = -i\hbar \int \Psi^{*}\frac{\partial \Psi}{\partial x} dx

Taking the derivative with respect to time gives

<br /> \frac{d \left&lt; p \right&gt;}{dt} = -i\hbar \int \frac{\partial}{\partial t}\left( \Psi^{*}\frac{\partial \Psi}{\partial x}\right) dx= -i\hbar \int \left[ \frac{\partial \Psi^{*}}{\partial t}\frac{\partial \Psi}{\partial x}+\Psi^{*}\frac{\partial}{\partial t}\left(\frac{\partial \Psi}{\partial x}\right) \right] dx = -i\hbar \int \left[ \frac{\partial \Psi^{*}}{\partial t}\frac{\partial \Psi}{\partial x}+\Psi^{*}\frac{\partial}{\partial x}\left(\frac{\partial \Psi}{\partial t}\right) \right] dx<br />

From Schrodinger's Equation,

\frac{\partial \Psi}{\partial t} = \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi

and

\frac{\partial \Psi^{*}}{\partial t}=-\frac{i \hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}} + \frac{i}{\hbar}V\Psi^{*}

Substitution gives,

\frac{d \left&lt; p \right&gt;}{dt} = -i\hbar \int \left[ \left( -\frac{i \hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}} + \frac{i}{\hbar}V\Psi^{*}\right) \frac{\partial \Psi}{\partial x}+\Psi^{*}\frac{\partial}{\partial x}\left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi\right) \right] dx

\frac{d \left&lt; p \right&gt;}{dt}= -i\hbar\int \left( -\frac{i \hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}} + \frac{i}{\hbar}V\Psi^{*}\right) \frac{\partial \Psi}{\partial x} dx - i\hbar \int \Psi^{*}\frac{\partial}{\partial x}\left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi\right) dx

Go to next post...

 

[buffordboy23]

Let

f = \Psi^{*} \Rightarrow \frac{df}{dx} = \frac{\partial \Psi^{*}}{\partial x}dx

and let

\frac{dg}{dx} = \frac{\partial}{\partial x}\left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi\right)dx \Rightarrow g = \left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi\right)

Using the product rule, then we have

<br /> \frac{d \left&lt; p \right&gt;}{dt}= -i\hbar\int \left( -\frac{i \hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}} + \frac{i}{\hbar}V\Psi^{*}\right) \frac{\partial \Psi}{\partial x} dx + i\hbar \int \frac{\partial \Psi^{*}}{\partial x}\left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi\right) dx -i\hbar\left\left[\Psi^{*}\left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi\right)\right] \right|^{+\infty}_{-\infty}<br />

Since \Psi^{*} goes to zero at positive and minus infinity

-i\hbar\left\left[\Psi^{*}\left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi\right)\right] \right|^{+\infty}_{-\infty} = 0

and now by taking the minus sign into the first integral

\frac{d \left&lt; p \right&gt;}{dt}= i\hbar\int \left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}} - \frac{i}{\hbar}V\Psi^{*}\right) \frac{\partial \Psi}{\partial x} dx + i\hbar \int \frac{\partial \Psi^{*}}{\partial x}\left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi\right) dx

go to next post...

 

[buffordboy23]

\frac{d \left&lt; p \right&gt;}{dt}= i\hbar\int \frac{i \hbar}{2m} \left(\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial \Psi}{\partial x}+\frac{\partial \Psi^{*}}{\partial x}\frac{\partial^{2}\Psi}{\partial x^{2}}\right)dx - i\hbar \int \frac{i}{\hbar}\left(\Psi^{*}V\frac{\partial \Psi}{\partial x}+\frac{\partial \Psi^{*}}{\partial x}V\Psi \right) dx

\frac{d \left&lt; p \right&gt;}{dt}= i\hbar\int \frac{i \hbar}{2m}\frac{\partial}{\partial x}\left(\frac{\partial \Psi^{*}}{\partial x}\frac{\partial \Psi}{\partial x}\right)dx - i\hbar \int \frac{i}{\hbar}\frac{\partial}{\partial x}\left(\Psi^{*}V\Psi\right) dx + i\hbar \int \frac{i}{\hbar}\left(\Psi^{*}\frac{\partial V}{\partial x}\Psi\right) dx

Now for the first two integrals let

f = constant \Rightarrow \frac{df}{dx} = 0

and let

<br /> \frac{dg}{dx} = \frac{\partial}{\partial x}\left(term\right)dx \Rightarrow g = \left( term\right)<br />

which will give us only a boundary term that equals 0 for the first two integrals from similar arguments from before. So,

\frac{d \left&lt; p \right&gt;}{dt}= \int \Psi^{*}\left(-\frac{\partial V}{\partial x}\right)\Psi dx=\left&lt;-\frac{\partial V }{\partial x}\right&gt;

Any problems this time around? =)

 

[buffordboy23]

It's called Clairaut's theorem or Young's theorem or something. There are mathematically pathological cases in which they don't. But they aren't of interest in physics. Assume partial derivatives with respect to the independent variables commute.

Yes, I was trying to determine this earlier and I came across the same theorem. It initially caused me to not want to swap derivatives, even when I could not come up with a counter-example, until I saw your remarks. Does mathematics sometimes seem too pure for nature? =) I do wonder what the disk D from this theorem would look like for points x and t; it would probably be extremely eccentric on an xt-plane since it makes sense the radius does not exceed x=ct.

 

[Phymath]

thats a lot of work when you can just do the general Ehrenfest Thrm

for O some operator
<br /> <br /> \frac{d &lt;O&gt;}{dt} = \frac{\partial}{\partial t} &lt;\psi|O|\psi&gt; = &lt;\dot{\psi}|O|\psi&gt; + &lt;\psi|O|\dot{\psi}&gt;+&lt;\psi|\dot{O}|\psi&gt;

<br /> -i |\dot{\psi}&gt; = H |\psi&gt;

<br /> \frac{d &lt;O&gt;}{dt} = i &lt;[H,O]&gt; + &lt;\frac{\partial O}{\partial t}&gt;

use p for O you get the same result in half the time...

 

[buffordboy23]

Phymath,

Nice presentation. I have yet to learn the notation that you used, but I see your arguments pretty clearly. I saw this same notation in my book but haven't got there yet. I initially figured that the formal notation would compact the equations into nice simplified forms for manipulation. I suppose that suffering is part of the journey towards learning QM well.

 

[dotman]

I wanted to bump this because I too ran into issues with this derivation. This is from problem 1.7 in Griffiths QM. I believe I've found a solution, but I'm not sure if I cheated on one critical step.

Essentially, my solution tracked with the OPs up until this point (the last eqn in his first full derivation post):

\frac{d \left&lt; p \right&gt;}{dt}= -i\hbar\int \left( -\frac{i \hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}} + \frac{i}{\hbar}V\Psi^{*}\right) \frac{\partial \Psi}{\partial x} dx - i\hbar \int \Psi^{*}\frac{\partial}{\partial x}\left( \frac{i \hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} - \frac{i}{\hbar}V\Psi\right) dx

At this point, I diverged and actually evaluated the partial derivative in the second integral:

\frac{d \left&lt; p \right&gt;}{dt}= -i\hbar\int \left( -\frac{i \hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}} + \frac{i}{\hbar}V\Psi^{*}\right) \frac{\partial \Psi}{\partial x} dx - i\hbar \int \Psi^{*}\left( \frac{i \hbar}{2m}\frac{\partial^{3}\Psi}{\partial x^{3}} - \frac{i}{\hbar}\left[\frac{\partial V}{\partial x}\Psi + V\frac{\partial\Psi}{\partial x}\right]\right) dx

I expanded and combined the integrals (actually, in my version, I had never split it into two integrals, but that's beside the point):

\frac{d\left&lt;p\right&gt;}{dt}=-i\hbar\int\left(-\frac{i\hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}+\frac{i}{\hbar}V\Psi^{*}\frac{\partial\Psi}{\partial x}\right)dx+\left(\frac{i\hbar}{2m}\frac{\partial^{3}\Psi}{\partial x^{3}}\Psi^{*}-\frac{i}{\hbar}\frac{\partial V}{\partial x}\Psi^{*}\Psi-\frac{i}{\hbar}V\frac{\partial\Psi}{\partial x}\Psi^{*}\right)dx

Next I collected like terms:

\frac{d\left&lt;p\right&gt;}{dt}=-i\hbar\int\frac{i\hbar}{2m}\left(\frac{\partial^{3}\Psi}{\partial x^{3}}\Psi^{*}-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}\right)dx+\frac{i}{\hbar}\left(V\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial V}{\partial x}\Psi^{*}\Psi-V\frac{\partial\Psi}{\partial x}\Psi^{*}\right)dx

And at this point I got stuck. I could see that the right term had my answer (Griffiths gives it to you as:

\frac{d\left&lt;p\right&gt;}{dt}=\left&lt;-\frac{\partial V}{\partial x}\right&gt;

If only the left term cancelled. But I didn't see how to do it. After taking a look at the first few steps of the OPs solution, I realized I could integrate the two parts of my first term by parts... not sure why this didn't occur to me before, I was too busy going back to see if I had made some mistake, as I was certain they should subtract out, or some such thing. So I split it up:

\frac{d\left&lt;p\right&gt;}{dt}=\frac{\hbar^{2}}{2m}\int\frac{\partial^{3}\Psi}{\partial x^{3}}\Psi^{*}dx-\frac{\hbar^{2}}{2m}\int\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}dx+\int\left(V\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial V}{\partial x}\Psi^{*}\Psi-V\frac{\partial\Psi}{\partial x}\Psi^{*}\right)dx

And integrate the first term by parts using:

u = \psi^{*} , dv = \frac{\partial^{3}\Psi}{\partial x^{3}} dx

du = \frac{\partial\Psi^{*}}{\partial x} dx , v = \frac{\partial^{2}\Psi}{\partial x^{2}}

Which gives me:

\frac{d\left&lt;p\right&gt;}{dt}=\frac{\hbar^{2}}{2m}\left(\psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}|_{-\infty}^{\infty}-\int\frac{\partial^{2}\Psi}{\partial x^{2}}\frac{\partial\Psi^{*}}{\partial x}dx\right)-\frac{\hbar^{2}}{2m}\int\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}dx+\int\left(V\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial V}{\partial x}\Psi^{*}\Psi-V\frac{\partial\Psi}{\partial x}\Psi^{*}\right)dx

And I evaluate the first portion of that integration by parts to go to zero, on the basis that \psi^{*} goes to zero at +/- infinity, leaving me with just:

\frac{d\left&lt;p\right&gt;}{dt}=-\frac{\hbar^{2}}{2m}\int\frac{\partial^{2}\Psi}{\partial x^{2}}\frac{\partial\Psi^{*}}{\partial x}dx-\frac{\hbar^{2}}{2m}\int\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}dx+\int\left(V\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial V}{\partial x}\Psi^{*}\Psi-V\frac{\partial\Psi}{\partial x}\Psi^{*}\right)dx

Where I can combine the first two integrals:

\frac{d\left&lt;p\right&gt;}{dt}=-\frac{\hbar^{2}}{2m}\int\frac{\partial^{2}\Psi}{\partial x^{2}}\frac{\partial\Psi^{*}}{\partial x}+\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}dx+\int\left(V\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial V}{\partial x}\Psi^{*}\Psi-V\frac{\partial\Psi}{\partial x}\Psi^{*}\right)dx

And simplify the first integrand to:

\frac{d\left&lt;p\right&gt;}{dt}=-\frac{\hbar^{2}}{2m}\int\frac{\partial}{\partial x}\left(\frac{\partial\Psi}{\partial x}\frac{\partial\Psi^{*}}{\partial x}\right)dx+\int\left(V\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial V}{\partial x}\Psi^{*}\Psi-V\frac{\partial\Psi}{\partial x}\Psi^{*}\right)dx

Now I evaluate the first integral to be:

-\frac{\hbar^{2}}{2m}\int\frac{\partial}{\partial x}\left(\frac{\partial\Psi}{\partial x}\frac{\partial\Psi^{*}}{\partial x}\right)dx=-\frac{\hbar^{2}}{2m}\left[\frac{\partial\Psi}{\partial x}\frac{\partial\Psi^{*}}{\partial x}\right]_{-\infty}^{\infty}

And here I say, and this is the part that I'm worried about, that since \Psi goes to zero at +/- infinity, then \frac{\partial\Psi}{\partial x} also goes to zero at +/- infinity, and thus this term is zero.

If this flies, it leaves me with the second portion, which reduces to the correct answer. But I don't feel 100% comfortable with my argument in the paragraph above.

Have I made a mistake with this argument? If not, how can I be sure I haven't (mathematically)? Thanks!

 

[makasx]

since \Psi goes to zero at +/- infinity, then \frac{\partial\Psi}{\partial x} also goes to zero at +/- infinity,

that helps me thx