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While there were already many questions on superposition in quantum mechanics, with great answers, these somehow leaves me uncertain about its physical meaning once I think about it a bit further.

Let me explain why the explanation with the double-slit experiment does not quite satisfy me. The system with the first (respectively, the second) slit open can be described by a Schroedinger equation of the form $$ ih\frac{\partial \Psi}{\partial t} = (-\Delta + V_{1,2})\Psi, $$ where the potential $V_{1}$ (resp. $V_2$) is, say, large negative at the first slit (resp., the second one) and zero elsewhere at the wall. If both slits are open, the equation can be written as

$$ ih\frac{\partial \Psi}{\partial t} = (-\Delta + V_{1}+V_2)\Psi. $$ If $\Psi_{1,2}$ are solutions to the former equation, then $\Psi_1+\Psi_2$ is approximately a solution to latter one. Note that we are actually adding states corresponding to two different physical systems, to obtain a state of a third system.

The same applies to the reflection example. We are using two different solutions to the free particle Schroedinger equation to cook up a solution to the equation with a barrier.

So far it all looks like merely a mathematical trick, exploiting nice features of the Schroedinger equation. We are not adding two states of the same physical system. So, the question is:

  • What is a physical meaning of adding two states of the same system?

One thing that concerns me here is the following. As far as I understand, the state space of a quantum-mechanical system is not, actually, a Hilbert space, but a complex projective space over a Hilbert space: the wave function should be normalized, and multiplication by $e^{i\alpha}$ leads to a physically indistinguishable state. Now, if quantum superposition had a physical meaning, then adding two states should be well-defined over this projective space, which it is not. In other words, $$ \frac{1}{\sqrt{2}} (|\psi_1\rangle+e^{i\alpha}|\psi_2\rangle) $$ corresponds to physically different states for different choice of $\alpha$, while $e^{i\alpha}|\psi_2\rangle$ describes the same state.

Kostya_I
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  • if you take your state $\frac{1}{\sqrt{2}} (|\psi_1\rangle+e^{i\alpha}|\psi_2\rangle)$ and multiply it by $e^{i\alpha}$ you DO get a physically indistinguishable state. Multiplication of a single factor like $\vert\psi_2\vert$ by a phase is not the same as multiplication of a full state by a phase. – ZeroTheHero Aug 03 '17 at 18:09
  • I think that @ZeroTheHero might have read your final paragraph too hastily and missed the point. You are right that even when $B$ and $B'$ represent the same state, $\gamma(A+B)$ and $\gamma(A+B')$ can represent different states, and, contrary to the apparent meaning of the above comment, no multiplication by a scalar can change that. – WillO Aug 04 '17 at 03:55
  • @WillO I removed my answer as I would rather not mislead the OP. I will re-read the qustion again. I agree $\gamma(A+B')$ will be different from $\gamma(A+B)$ and I thought this is exactly what my comment expressed... – ZeroTheHero Aug 04 '17 at 07:34
  • @zerothehero: it's of course possible that your answer was fine and I misread it; I can't doublecheck now that you've deleted. The content of the answer was certainly fine, but I did think it was answering a different question than the one that was asked. – WillO Aug 04 '17 at 12:26
  • @WillO No worries. I'll double check later today. – ZeroTheHero Aug 04 '17 at 12:48

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