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So now that the quiz is over I think it's fair game to discuss the pointless subject of <math><\delta p/\delta t></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, but the partial of p should be zero. Is that right, and if so can someone explain why?

Also, did you type that all up in LaTeX?

Yeah, I wrote it all in LaTeX. Making pretty things is fun. Theorem 3.71 asks for an explicit time dependence, so I think <math><\delta p/\delta t>=0</math> (I fixed all the derivatives to be partial - sorry for the typo. All derivatives inside the expected value should be partial). That's why I say it's actually pointless.

My understanding of why the <math><dp/dt></math> is not equal to <math><\partial p/\partial t></math> is that these two things has totally different meanings mathematics. <math><dp/dt></math> is the time deriavate on the operator <math>\hat{P}</math> , in the other words a function of <math>\hat{P}</math>. But <math><\partial p/\partial t></math> is explained as the product of two operators <math>\hat{H}\ and\ \hat{p}</math>. Since <math>\hat{H}</math> and <math>\hat{p}</math> are all Hermitian operators, they commute with each other.

I'm really confused by what Griffiths is doing on the bottom of page 103. What is he trying to accomplish with the integral in Equation 3.31? Is that the inner product of different eigenfunctions for the momentum operator? Also, how does he get a Dirac-delta function for the result of that integrap?

The integral in Equation 3.31 is <fp'|fp> (like you said, the inner product of two eigenfunctions for the momentum operator).

Correct me if I'm wrong, but I think the Dirac-delta function comes from the fact that the integral is zero unless p=p'. The way I think about it is to rewrite the exponential in terms of cosine and i*sine–the integral of these two from -∞ to +∞ is zero, so unless p=p' (making the integrand 1), the integral must be zero. This is what Griffiths calls “Dirac orthonormality”–it's not true orthonormality because p can also have complex values, which is not taken into consideration in this calculation (and true orthonormality has a Kronecker delta, not a Dirac delta).

It is helpful when you see page 70.<math>f(x)\delta(x-a)=f(a)\delta(x-a)</math>,(because the product is zero anyway except at the point a.) Furthermore, using Eq.2.113 <math>\int f(x)\delta(x-a)dx=f(a)\int\delta(x-a)dx=f(a)</math>,you can get Eq. 3.35.

In the note at the bottom of page 104 Griffiths writes, “any complex number is an eigenvalue of the operator <math>\hat{p}</math>, but only real numbers are eigenvalues of the hermitian operator <math>\hat{p}</math>. Can anyone paint a picture of what he means here.

My understanding is the hermitian operators have a special condition, just like the definition of 3.16 on p97. In order to satisfy this condition, the eigenvalues of the momentum operator should be real numbers. If eigenvalues or the momentum operator are not real number, this definition is not satisfied. In other words, the imaginary part of an eigenvalue will be changed.

For an operator to be Hermitian, all the eigenvalues must be real (and conversely, if all the eigenvalues are real, the operator is Hermitian, or at least I think that's what Yuichi said in class). Therefore, if we consider <math>\hat{p}<math> a Hermitian operator, we can only consider its real eigenvalues. <math>\hat{p}<math> also has complex eigenvalues, but with these values the eigenfunctions are non-normalizable and blow up at ±∞ (i.e., they do not exist in Hilbert space), meaning that the operator cannot be Hermitian.

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