# Schrödinger theory, bunch of questions

1. Nov 1, 2013

### fede.na

Hi, I've got a few questions about Schrödinger's formulation of QM, mostly about how to interpret the results:

1) How do I choose boundary conditions? I know that I should always normalize the wavefunction and make it continuous but some times you need to make the derivative continuous as well. I get a feeling that the first derivative should always be continuous since there is a Laplacian in the equation, otherwise it would be infinite. But if I remember correctly, the solution of the dirac's delta well does not have a continuous derivative.

2) What do I get when I apply an operator? let's say there is a magnitude $f(\textbf{r},t)$then $\hat{f}\Psi = f(\textbf{r},t)\Psi$. Now, this object $f(\textbf{r},t)\Psi$ that somehow resembles a wavefunction, what is it? I know that doing this returns an expected value for the magnitude, in one dimention $<f>= \int_{-\infty}^{+\infty} \Psi^*f(x,t)\Psi dx$. But, without integrating, is there anything I can know from the result of the operator $f(\textbf{r},t)\Psi$. It looks as though it is the probability distribution of the magnitude f, but I'm not quite sure how to interpret that, since the distribution is in terms of x, and not of the values of the magnitude.

3) What role does the Fourier Transform play in all this? I gather that you can move from momentum to position, transforming the distributions, but I'm not sure how to do it. For any given wavefunction, can the momentum distribution be obtained from squaring $\int_{-\infty}^{+\infty}\Psi (x,t)e^{-i2\pi f x} dx$ ? If so, from the previous question, does this equal $\hat{p}\Psi = -i\hbar\frac{\partial}{\partial x} \Psi$ ? Again, I get the feeling this can't be right just because the variables don't make sense, the Fourier transform will not be a function of x.

4) What role do initial conditions play in this theory? Since it is a differential equation, initial conditions always tend to change the problem a lot. Other that boundary conditions, how are initial conditions in time taken into account? How do you measure them without destroying the system? Or do you just assume every possible solution at the same time until the system is observed?

5) In the finite potential barrier problem, when the particle has less energy than the barrier, tunneling occurs. As weird as tunneling is, I don't really have a problem with it. What is bugging me is the fact that there is a non-zero probability of finding the particle inside the barrier. Never mind how small it is, it's not zero. Wouldn't that violate the conservation of energy? Say you have a particle emitter that emits particles of a certain ammount of energy, within a certain uncertainty. Say the barrier is 10 or 100 times bigger than that, there is still a chance of finding particles within the barrier, but that would imply they have at least the potential energy of the barrier, which is absurd because we know it started with way less than that... you get the idea

6) Finally, kind of a big question that perhaps should go in a separate thread, but I don't want to spam the forum with newbie questions :). Why aren't the solutions to the Hydrogen atom symmetric across the angular coordinates? In electromagnetism, when solving for a point charge or charged sphere, I loved the fact that you could dispose of most of the Laplacian because it totally made sense: the problem is symmetric, there shouldn't be an angular dependency... there's no physical reason for it to happen. But the most compelling argument, as I see it, is this: How does the atom know which directions I chose in my coordinate system? The problem is symmetric so there is no way of relating my particular system of coordinates to a set of directions in an actual physical system, because there aren't any "irregularities" to let you distinguish a particular direction from the other ones, they are all in equal standing.

I know they are a ton of questions but this is a really hard topic! Thank you a LOT in advance

2. Nov 1, 2013

### Staff: Mentor

For your question #6: the atom "knows" and the angular dependency appears because you're choosing to measure the angular momentum about a particular axis, and this is more than just a choice of coordinates as it is in classical physics. A measurement alters the state of a quantum system, and a choice to measure the component of angular momentum on one particular axis instead of settling for the conveniently symmetrical $L^2$ introduces an asymmetry into the system.

You may object that you're not measuring anything, you're just solving a differential equation that describes the state of the atom. But that's not quite right - you're solving a differential equation that predicts the results of measurements made on the atom. There's no useful distinction between the two in classical mechanics, but in quantum mechanics there is.

3. Nov 2, 2013

### Jano L.

1) Continuous/discontinuous depends on you, sometimes either may be useful. In addition, also $\psi \rightarrow 0$ at infinity is useful.

2) the result has generally no direct physical meaning, the meaningful thing is $\int \psi^*f\psi dx$.

3) One can derive new function $\tilde{\psi}(p)$ from $\psi(x)$ by a customized version of the Fourier transfom. This function has the property

$$\int \psi^*(x) \hat{p} \psi(x) dx = \int |\tilde{\psi}(p)|^2 p dp,$$

so $|\tilde{\psi}(p)|^2$ is sometimes considered to be probability density of momentum.

4) For initial condition, often the first eigenfunction of the Hamiltonian is assumed, or equilibrium statistical mixture of them, each with weight $e^{-\frac{E}{k_B T}}$*from the Boltzmann distribution (most often density matrix is used for that). $\psi(x)$ cannot be measured in atoms; except for spin function, which can be prepared by magnets.

5) Yes, there is energy fluctuation, if we assume that the particle actually goes there. Alternatively people invented smart elusions to this, like to say that if you do not measure it, its energy does not exist, and if you measure it, it you who supplies the extra energy.

6) The angular dependence is because what actually Schroedinger did was finding eigenfunctions of spherical Laplacian. This is most easily done in spherical polar coordinate system, which has certain special directions, like those of the axis and of the $\varphi = 0$ meridian. The orientation of these is arbitrary, no prediction depends on it in the end, but the intermediate steps do, like the eigenfunctions $\Phi_{nlm}$ chosen to expand the psi function $\psi(\mathbf r, t)$. It is the same thing that happens in continuum mechanics of Earth or a star or anything spherical; the modes have to be chosen somehow, and the easiest way to do it is via polar coordinates.

I wonder whether there is some other system of spherical coordinates that does not have this asymmetry...