Free particle in spherical coordinates

Question

I'm trying to solve the very simple equation:

$$-\frac{\hbar^2}{2m}\nabla^2 \psi = E\psi$$

but in polar coordinates. I used separation of variables to find out that my wave function is of the form: $$\psi = R(r)Y(\theta, \phi)$$

as expected, and $R(r)$ satisfies the radial part of the differential equation. However, I noticed I could solve it in cartesian coordinates, then turn into polar coordinates:

$$\psi(x,y,z) = e^{i (k_x x + k_y y + k_z z)} = e^{i(k_r r + r^2 k_\theta \theta + r^2 \sin^2(\theta) k_\phi \phi)}$$

with appropriate units. Now why are these two wave functions different?

Also, I tried applying the radial momentum operator on $\psi$ and in the second representation, I don't get what I expect:

$$p_r \psi = -i\left(\frac{\partial}{\partial r } - \frac{1}{r} \right)\psi \neq k_r \psi$$

which I would expect. What is wrong here?

Answer 1

The answer is what Javier said, you have found two solutions to the differential equation in different coordinates, and the general solution will be a superposition of all solutions in some coordinates, made into a specific solution by initial or boundary conditions. Think about the regular Laplace problem: $\nabla^2\phi=0$, one solution (stricly speaking for $r\neq0$) is $\phi=1/r$, another is $\phi=x+y+z$, and it is very clear that $x+y+z\neq\frac{1}{r}$, but that's ok because partial differential equations have many very different looking solutions unless you have boundary conditions. Whether using spherical or cartesian coordinates is more appropriate will depend on the situation at hand. If its plane waves scattering off a localized disturbance you would use both (incoming monochromatic plane wave, outgoing spherical waves). Also, for the radial momentum it should be a plus sign, not a minus sign.

Answer 2

If you have studied about the hydrogen atom, you can immediately write the solutions :-

$Y(\theta, \phi) = \sqrt{\frac{2l + 1}{4\pi}\frac{(l-m)!}{(l+m)!}} P^{m}_{l}(\cos{\theta}) e^{i m\phi}$

$R(r) = Aj_{l}(kr) + Bn_{l}(kr)$

I hope you know about all the functions and quantum numbers.

The solution that you have found is not wrong, it is a solution of particle in free space expressed in polar coordinates, but it is not the solution of the time-independent schrödinger equation. What it means is that your solution is not an eigenfunction of the hamoltonian, it is a "sum"(superposition, to be more precise) of the eigenfunctions of the hamoltonian, which are $Y(\theta,\phi)R(r)$, for integer values of $m$ and $l$. This is the reason of getting non equal values of the momentum. And yes, mike is right, there should be a plus sign in front of $1/r$ in the very last equation.

Answer 3

Remember that in $\mathbb R^n$, there is a continuum of solutions to the Schrödinger equation for each energy $E$: $He^{ik\cdot x} = \frac{\hbar^2 k^2}{2m}e^{ik\cdot x} = Ee^{ik\cdot x}$. In principle, any linear combination of these elementary solutions could have been used as an eigenfunction, and these linear combinations will tend to look very different from simple plane waves. The exact amount of 'wiggle room' this affords in the choice of basis wave functions is a deep mathematical question that you might say is at the heart of harmonic analysis (to start, you might consider the extent to which it's possible to infer results about harmonics of a given frequency over non-compact spaces from harmonics over compact subspaces, using basic results from Fourier analysis, which incidentally might also help with computing the desired harmonics in polar coordinates.)

As a hint for why your transformation failed to produce an eigenfunction (of the Laplacian, or $\partial_r$), consider that the wave function is a special type of distribution, whose squared norm is interpreted as a probability density function (how do probability density functions transform under coordinate transformations?) Also, $\partial_r$ is a very different operator from $\partial_x$, $\partial_y$, and $\partial_z$. Think about how $\partial_r$ acts on functions in small neighborhoods of $\mathbb R^n$: on the positive $x$ axis its effect resembles $\partial_x$, but then it rotates to $\partial_y$ on the positive $y$ axis. Even locally, except maybe on the positive $x$ axis, you shouldn't expect an eigenfunction of $\partial_x$ to look like an eigenfunction of $\partial_r$ (and vice versa.)