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Is it accurate to say that the solutions to the Schrodinger Equation tell us the allowed outcomes of a quantum experiment and the probability of observing that outcome?

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Is it accurate to say that the solutions to the Schrodinger Equation tell us the allowed outcomes of a quantum experiment and the probability of observing that outcome?

"Quantum experiment" is a very general statement, covering all the reactions in the microcosm.

The Schrodinger equation is one of the quantum mechanical equations where two body potential problem can be explicitly solved. The most successful is the solution of the energy levels of Hydrogen and hydrogen like atoms using the Schrodinger equation.. The other equations are the Dirac, for fermions, the Klein Gordon for bosons and the quantized Maxwell equation for photons. These cover the interactions of all the particles in the standard model of particle physics.

If more than two particles in a potential well are involved, approximations to possible potentials might give a solution describing the data, depending on the experiment.

Fortunately theory has progressed to second quantization, which uses the free ( no potential) solutions of the appropriate equations to define a field for each elementary particle, on which creation and annihilation operators allow for a more complex calculation that can give the probability of interaction. This is quantum field theory.. The calculations involve the evaluation of Feynman diagrams.

So it is more complicated than a simple potential solution to get at the probabilities.

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No. QM works like this:

  1. The universe occupies some initial state, usually described by a wavefunction $|\psi_0\rangle,$ in some abstract space of all possible states.
  2. The things we want to measure are described with Hermitian linear operators $\hat O^\dagger = \hat O.$ The possible outcomes must be eigenvalues of the operator, $\hat O|o\rangle = o |o\rangle,$ and quantum mechanics will only predict averages over many experiments, $\langle O \rangle = \langle \psi|\hat O |\psi\rangle.$ This also gives you some visibility on how to describe the earlier wavefunction.
  3. We make sure that the operators have various commutation relations to put the Heisenberg uncertainty principle into practice. Some of these are very well-known recipes; sometimes it takes a while to see how to do this step.
  4. We identify one of our above linear operators with the total energy of the system, $\hat H.$
  5. We now implement the Einstein-Planck relation $E = h f$ via the Schrödinger equation $i\hbar \partial_t |\psi(t)\rangle = \hat H |\psi(t)\rangle,$ with $|\psi(0)\rangle = |\psi_0\rangle.$
  6. Then we use the resulting function to predict those expectation values for the above operators, and fiddle with our definitions of operators and their commutation relations until we match with experiments.

Notice that the "allowed outcomes" are decided somewhere in parts 1, 2, and 3; the Schrodinger equation just comes out of these decisions and tells us what wavefunction we've got and therefore what all of the predictions are among those allowed outcomes.

To give a concrete example, solving the Schrödinger equation for the Hydrogen atom does not tell us anything about where the electron is allowed to be; it's allowed to be anywhere in space or in any superposition of anywheres in space. But it does tell us how that $|\psi_0\rangle$ changes over time and this of course affects all of our experimental results.

Another concrete example: in the double-slit experiment we see photons impinge upon a detector located at a vertical position $y,$ the "observable" here is whether this detector "ticks" or "doesn't tick", call these $|\tau_y\rangle$ and $|\nu_y\rangle$ respectively. The Hamiltonian says that the wavefunction for a particle that moves through the first slit to make the detector tick is some $|\sigma_1(y)\rangle = g_1(y) |\nu_y\rangle + f_1(y) |\tau_y\rangle,$ and for a particle which moves through the second slit it is instead some $|\sigma_2(y)\rangle = g_2(y) |\nu_y\rangle + f_2(y) |\tau_y\rangle.$

Our observable is that we observe ticks at $y$; their average number will be given by the Hermitian observable $\hat T_y = |\tau_y\rangle\langle\tau_y|.$ Therefore if we just open up slit 1 and close slit 2 we see $$\langle T_y\rangle = \Big(g_1^*(y) \langle\nu_y| + f_1^*(y) \langle\tau_y| \Big) |\tau_y\rangle\langle\tau_y| \Big(|\sigma_1(y)\rangle = g_1(y) |\nu_y\rangle + f_1(y) |\tau_y\rangle\Big),$$ reducing all the way down to: $$\langle T_y\rangle = f_1^*(y) f_1(y) = |f_1(y)|^2.$$It's $|f_2(y)|^2$ for just opening slit 2 with slit 1 closed. Now we open both and put the system into this state $\sqrt{\frac12} |\sigma_1(y)\rangle + \sqrt{\frac12} |\sigma_2(y)\rangle$, what happens? We instead get, once everything resolves, $$\langle T_y\rangle = \frac12 | f_1(y) + f_2(y) |^2.$$ And now if you think that say $$f_k(y) = e^{2\pi i (y-a_k)/\lambda} ~e^{-(y - a_k)^2/(4 b^2)},$$ then we see in the one-slit case a bell curve of light of width $b$ centered at $a_1$ or $a_2$. But when we open both slits, we see an interference pattern, as long as $b$ is big enough so that the bell curves overlap, while $\lambda$ is small enough that we get to see a few wavelengths of light between the two centers. So there is "no interference pattern" for one slit open (either slit), but there is an interference pattern for both slits open.

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  • $\begingroup$ Isn't it true that the quantum numbers for hydrogen follow directly from the solution of the Schrodinger equation. Here is a quote "Solution of these equations under the constraints placed on the wavefunction leads to series solutions in the form of polynomials called the associated Laguerre functions. In order to fit the physical boundary conditions, these solutions contain a parameter n which can take only positive integer values; this parameter is called the principal quantum number." $\endgroup$ – hooch Feb 5 '17 at 18:35
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    $\begingroup$ @hooch: Yes. However be very careful here because you might naively think that this contradicts anything I've said above, and it doesn't: the spherical harmonics do in fact span the vector space of $|\psi\rangle$, but so do the position-basis states. So any wavefunction has two equivalent descriptions, one in terms of "here is how it is distributed over space' and another in terms of "here is how it is distributed over energy and angular momentums." The allowed outcomes have not changed: we have merely found a better way to describe those outcomes, e.g. in the limit of low internal energies. $\endgroup$ – CR Drost Feb 5 '17 at 18:50
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    $\begingroup$ Discrete or continuous in what space? Do you think a plane wave is continuous (position space) or discrete (momentum space)? $\endgroup$ – CR Drost Feb 5 '17 at 19:03
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    $\begingroup$ No, I don't agree to that: one of the allowed states is the ground state of the Hydrogen atom (the s-orbital) and it is continuous over position-space. $\endgroup$ – CR Drost Feb 5 '17 at 19:16
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    $\begingroup$ @Emil right, my bigger point here is that "allowed states" is not a very useful idea in QM, instead you have "what are the allowed outcomes of this or that measurement"? as the closest helpful thing you can ask. $\endgroup$ – CR Drost Feb 5 '17 at 19:19

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