Return to Q&A main page: Q_A
Q&A for the previous lecture: Q_A_1023
Q&A for the next lecture: Q_A_1028

If you want to see lecture notes, click lec_notes

Main class wiki page: home

So now that the quiz is over I think it's fair game to discuss the pointless subject of <math><dp/dt></math>. So if it is in fact equal to <math>←i\hbar\frac{\delta^2}{\delta t\delta x}></math>, let's add 0 to this by adding the right side of equation 1.38 and subtracting the left side, substituting in the definition of momentum. Then, let's expand the definition of expectation, moving the time derivative inside the integral:

<math>←i\hbar\frac{\delta^2}{\delta t\delta x}>=←i\hbar\frac{\delta^2}{\delta t\delta x}>-\frac{d}{dt}←i\hbar\frac{\delta}{\delta x}>+←\frac{dV}{dx}>=←i\hbar\frac{\delta^2}{\delta t\delta x}>-\int_{-\infty}^\infty\frac{\delta}{\delta t}[\Psi^*(-i\hbar\frac{\delta}{\delta x}\Psi)]dx+←\frac{dV}{dx}>=</math>

Now, let's apply the product rule. Look, one of the resulting terms (the third term in the sum) is just the negative definition of the first term, the expectation of the operator we are talking about! They cancel. Now, let's integrate by parts. As usual, the boundary term vanishes because wavefunctions go to zero at <math>\pm\infty</math>

<math>=←i\hbar\frac{\delta^2}{\delta t\delta x}>-\int_{-\infty}^\infty\frac{\delta}{\delta t}\Psi^*(-i\hbar\frac{\delta}{\delta x}\Psi)dx-\int_{-\infty}^\infty\Psi^*(-i\hbar\frac{\delta^2}{\delta x\delta t}\Psi)dx+←\frac{dV}{dx}>=-\int_{-\infty}^\infty\frac{\delta}{\delta t}\Psi^*(-i\hbar\frac{\delta}{\delta x}\Psi)dx+←\frac{dV}{dx}>=\int_{-\infty}^\infty\frac{\delta^2}{\delta t\delta x}\Psi^*(-i\hbar\Psi)dx+←\frac{dV}{dx}>=</math>

Now, let's move the <math>-i\hbar</math> onto the first term of the product, and inside the complex conjugate. Notice that we have to reverse the sign on i when doing this. We also move the double derivative inside the complex conjugate.

<math>=-\int_{-\infty}^\infty(-i\hbar\frac{\delta^2}{\delta t\delta x}\Psi)^*\Psi dx+←\frac{dV}{dx}></math>

But hey! The first term of the overall sum is the definition of the complex conjugate of the operator we are talking about! (We have no reason to think it's Hermitian) So, we have:

<math>←i\hbar\frac{\delta^2}{\delta t\delta x}>+←i\hbar\frac{\delta^2}{\delta t\delta x}>^*=←\frac{dV}{dx}></math> or <math>←i\hbar\frac{\delta^2}{\delta t\delta x}>+←i\hbar\frac{\delta^2}{\delta t\delta x}>^*=\frac{d}{dt}←i\hbar\frac{\delta}{\delta x}></math>

Since the RHS of these equations in general is not 0, this operator is not 0. The question is, is it Hermitian? If it is not, then this is not observable, and this is all truly pointless. If it is, then we have the following, which is an interesting parallel to how the quantum velocity is half the classical velocity.

<math>2←i\hbar\frac{\delta^2}{\delta t\delta x}>=\frac{d}{dt}←i\hbar\frac{\delta}{\delta x}></math>

That's a really clever technique you've used. But is your initial assumption, which is that <math><dp/dt></math> is in fact equal to <math>←i\hbar\frac{\delta^2}{\delta t\delta x}></math>, correct? I think it is, but I'm not sure.

However, since the inspiration for this discussion is from homework problem 3.17, and the theorem in the book (3.71 on p.115), let me ask this: In problem 3.17, we're not looking for <math><dp/dt></math>. We're looking for <math><\partial p/\partial t></math>. Are they equal? If I understood my TA correctly, they're not. The first one is the one you worked above, the partial should be zero. Is that right, and if so can someone explain why?

Also, did you type that all up in LaTeX?

Return to Q&A main page: Q_A
Q&A for the previous lecture: Q_A_1023
Q&A for the next lecture: Q_A_1028