I have recently started studying quantum mechanics, and here are some things that are really confusing me.

  1. Particle in a box: Supposedly, the square of the magnitude of the normalized wave function gives the probability that the particle will be in a specific location. I understand that for the superposition states, the states add up so that the wave function of some particle that can be found in some superposition of eigenstates with various energies, such that the eigenstates have wave functions $\psi_1, \psi_2, ...$, then a particle whose initial wave function is of the form $\sqrt{p_1}\psi_1+\sqrt{p_2}\psi_2+...$ will be found in the eigenstate corresponding to $\psi_1$ with probability $p_1$, the eigenstate corresponding to $\psi_2$ with probability $p_2$, and so on. But, if you took the square of this superposition wavefunction in order to get a probability distribution, it won't, in fact, be the same as the simple superposition of the original probability distributions with weights according to the probabilities, because there are cross terms. The explanation I've heard for this is that the integral of each cross term over the width of the box is zero, so that the wave function can still be normalized. However, the cross terms can very much be positive at some locations and negative at others, so that the resulting probability distribution from squaring the wavefunction is different from what you'd get from making a superposition of the wave functions of the eigenstates according to their probabilities.

  2. Normalization of the hydrogen atom wave functions: I'm struggling to understand how the $r^2sin(\theta)$ term (which I conceptually understand represents the ratio of volume between a voxel in spherical coordinates and a voxel with the same dimensions in Cartesian coordinates. In particular, how does it affect normalization? For instance, if you already have a normalized radial wave function $R_{n,l}$ and a normalized spherical harmonic $Y_{l,m}$, it is said that the product $R_{n,l}\times Y_{l,m}$ will be a solution to the eigenstate wave equation $\psi_{n,l,m}$. That being said, since $R$ and $Y$ are both normalized, and there is no overlap in the variables upon which they depend, their product should also be normalized. Except, there is an $r^2sin(\theta)$ term in the integral for the product, since we are integrating over volume in spherical coordinates. The $sin^2$ was already accounted for in the normalization of the spherical harmonic, but, in light of that, would the normalized wave function simply be equal to $\frac{R_{n,l}\times Y_{l,m}}{R}$, in order to account for the fact that, if the particle has an equal probability of appearing at two different radii, and one is farther out than the other, the density w/ respect to volume will be lower at the farther radius (and to cancel out the $r^2$ term in the normalization factor)?


closed as too broad by DanielSank, John Rennie, GiorgioP, Kyle Kanos, Chris Apr 28 at 17:26

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    $\begingroup$ For future reference note that the rule is one distinct question per post. $\endgroup$ – StephenG Apr 28 at 3:07
  1. In fact, the integral of the cross-terms is zero because they are eigenvectors of the Hamiltonian, a hermitian operator. It is a very common result in linear algebra that eigenvectors of hermitian operators are orthogonal to eigenvectors with different eigenvalues.

But the question 1 was about probability distribution from squaring the wavefunction. You state that it would be different from the probability using superposition of eigenstates according to their probabilities.

The superposition that exists in Quantum Mechanics is related to the wave function; it's not a superposition "in probabilities". In fact, the expression $\sqrt p_1 \psi_1 + ... + \sqrt p_n \psi_n $ comes from the superposition principle. The state can be written as a superposition of eigenstates (because eigenstates of a Hermitian operator form a base to the space), with some constants. $\psi = c_1 \psi_1 + ... + c_n \psi_n $. Now, using the fact that $(\left < \psi_k | \psi \right >)^2$ (squared modulus of the intern product) should be the probability of $\psi$ be in $\psi_k$, we get that $|c_k|^2 = p_k$.

That's the same thing that happens with light; we don't sum light intensities when we deal with diffraction, for example. We sum the amplitudes multiplied by their phases (complex exponential), and then we square them to get the resulting intensity (which is not the sum of both original intensities, there is the interference (cross) term. This term gives us constructive or destructive interference), as in quantum mechanics. The coeficients $c_k$ are amplitudes of each eigenstate, but there are crossed terms when we square the wave function, giving us interference terms. These terms only become irrelevant when we get the integral of $|\psi|^2$ in the whole space.

  1. Our normalizations can be written (omitting indices, for simplicity)

$$\eqalign{ &\int d^3 \vec r\>|R \cdot Y|^2 = {}\cr &\qquad \int\!\!\int\!\!\int r^2 \sin\theta\,dr\,d\theta\,d\phi\>|R \cdot Y|^2 = 1 \cr}$$

How the normalization is "divided" between radial and angular function is not important. If we divide R by 2 and multiply Y by 2, it wouldn't chance anything physically, since the wave equation is the product of radial and angular parts. So, usually what we do is that we normalize them separatelly. But this has to be done including the $r^2 sin\theta$ in this normalization, since they appear in the normalization integral. So, usually, we get:

$$\int d^3 \vec r |R \cdot Y|^2 = \left( \int r^2 dr\>|R(r)|^2 \right ) \cdot \left(\int \sin\theta\,d\theta\,d\phi\> |Y(\theta, \phi)|^2 \right)$$

And we make both integrals equal to 1 separatelly. (as I said before, one could make the radial one equals 3, as long as the angular one gives us 1/3. That's no difference. What matters physically is their product. Taking both integrals equal 1 is just a convention)


Regarding question one, that is correct. Quantum mechanics is a probabilistic theory, but it is fundamentally different from theories of classical probability.

Not to beat this horse too much, but the simplest example of how things are just fundamentally different for a quantum system that I know of goes like this: Alice, Bob, and Carol are working together on a team, and we give them a bunch of challenges where we put them into completely isolated rooms and make them press buttons labeled 0 or 1 to control the sum of their numbers. 25% of the time, we ask them all to make the sum even, and they win if it's even. 75% of the time, we choose one at random to work at cross-purposes to the other two: we ask the "traitor" to make the sum even, and the other two to make the sum odd, and the team wins if it is odd. (The "traitor" is still part of the team; they also fail if the sum is even. We're not paying them money to betray their fellows; we're just feeding them a wrong goal and watching the mayhem unfold.) In classical probability, you have to choose one of the four circumstances to fail in, so your maximum win probability is 75%. In quantum probability, the only thing stopping you from winning 100% of the time is how delicate quantum states often are: if you can give each one of these people a quantum bit, and pre-arrange for them to be in a special "superposition" state that we call a "GHZ state", and they can keep those states isolated from the rest of the world and manipulate and measure it correctly, then they can win this game 100% of the time.

So indeed, yes, you need to throw that original idea out the window, and indeed this is a fundamental part of the double-slit experiment: a photon going through the first slit displays some bell curve on a detector $|f_1(z)|^2$, a photon going through the second slit displays some bell curve on a detector $|f_2(z)|^2$, but a photon which goes through both displays some wavy interference pattern, $$\frac12 |f_1(z) + f_2(z)|^2 \ne \frac12 |f_1(z)|^2 + \frac12 |f_2(z)|^2.$$But, what is very interesting is that any interaction of that photon with something which is altered differently when it goes through slit 1 versus slit 2, tends to make the pattern on the right come back again. This is that same decoherence problem in miniature. For example if you rotate the polarizations by $\pm 45^\circ$ so that they are 90° separated, you see two overlapped bell curves with no waviness.

Regarding question two, there are actually many different conventions for normalizing the spherical harmonics and your proposed convention (multiplying each one by $\sqrt{\sin\theta}$ so that you don't have to write it in the integral) is certainly a valid alternative choice. I think people just don't choose it because the term $\sqrt{\sin\theta}$ looks ugly and it's easy enough to remember that the orthogonality condition is (in one of those many different conventions) $$\int_0^\pi d\theta \int_0^{2\pi}\sin\theta~d\phi~Y^m_\ell(\theta, \phi)~Y^{m'}_{\ell'}(\theta,\phi) = \{1 \text{ if } m=m' \text{ and } \ell =\ell' \text{ else } 0\}.$$Part of this is that in a paper it is easiest to just write $\iint d\Omega$ or even just $\int d\Omega$ but everybody already agrees on $d\Omega$ as a little bit of "solid angle" $\sin\theta d\theta~d\phi$ and so laziness suggests leaving the $\sin\theta$ in the integral so that you can just write $d\Omega$ rather than the twice-as-long $d\theta~d\phi.$ But yeah, there are lots of conventions for normalization where some people replace that 1 with $4\pi$ or even the Racah normalized $4\pi/(2\ell + 1)$, and multiply the underlying functions accordingly; I can't see a reason why you wouldn't be allowed to multiply one of those by a $\sqrt{\sin\theta}$ if you really wanted to.

The moral of the story is that normalization of a part of a wavefunction does not directly have a formal meaning, and needs to be approached with some trepidation. Only the normalization of the full product of all of the components has a meaning, and it only has that meaning by virtue of being a probability.


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