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What is the difference between Dirac's delta function and the Kronecker delta function?

It seems to me they are related only in name. The Kronecker delta is just a function of two variables that is 1 if they are equal and 0 if they are not. The Dirac delta function is a mathmatical construct to help us solve some problems

To expand on Dark Helmet's answer, the Dirac delta function is a function that is zero everywhere except at δ(0), where it is infinite. The integral of a Dirac delta function is always 1 (provided that the integral includes δ(0)). The Kronecker delta is most useful when defining orthonormality–whenever m≠n, the inner product of two eigenfunctions is zero; whenever m=n, the inner product is 1 (provided the functions are normalized).

It seems to me that the difference is that Dirac's delta function is for continuous case while Kronecker delta function is for discrete case.

What is a determinate state?

As far as I can tell, determinate states are just eigenfunctions of certain operators. For example, the stationary states we've been studying are determinate states of the Hamiltonian. For the stationary states, every measurement of the particle in a stationary state gives a corresponding energy. It really is just eigenfunctions and eigenvalues.

What did y'all think of the test? I thought there was less time pressure, which meant I could take the time I needed to do the best solutions I could.

When developing the differentiability boundary for the delta potential, why does the integral -epsilon to epsilon when integrating E(Psi) go to 0?

Because the integral of any function at just one point is zero; it's the area of a line. Since you're taking the limit of the integral as epsilon goes to zero, the integral of psi goes to zero.

Unless it's the magic delta function, of course. Which is part of what makes the delta function so useful, and also what makes it feel like math legerdemain*.

* sorry, too many GRE flash cards. I meant math magic.

Griffiths says “degenerate” is two or more linearly independent eigenfunctions share the same eigenvalue. Could you explain more about this with physical concepts or examples? Thanks,

Think of it like degenerate energy levels for a particle. Different configurations of elections (eigenfunctions) can cause the particle to have to same energy (eigenvalue).

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, 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 second term 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>

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