This is a question parallel to this question The importance of the phase in quantum mechanics.

In introductionary quantum mechanics I have always heared the mantra

The superposition principle which says that two states of a quantum system can be added to obtain a new state explains the interference we see in the double-slit experiment.

For example if after measuring a particle is equally likely be found in a state of spin up and spin down, its wave function is

$$ | \psi\rangle = \frac{1}{\sqrt{2}}| \text{up} \rangle + \frac{1}{\sqrt{2}} | \text{down} \rangle.$$

The coefficents squared give the probability of this particular pure state. Because of this the wave function of a quantum state always needs to be normalised. But then you can almost never form the superposition $| \psi_1 \rangle + | \psi_2 \rangle$ of two states, since the sum in general doesn't have norm one.

In fact, if we choose $| \psi_2 \rangle = - | \psi_1 \rangle$, the result vanishes. But I have often seen this as an explanation of the destructive interference we see in some quantum experiments: Two states combine (by their sum) to form constructive / destructive interference. How can this be made precise and what is the exact formulation of the superposition principle such that it makes mathematical sense? How could one model e.g. the double-slit experiment unsing this to describe constructive / destructive interference?

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    $\begingroup$ $(-1)^2=1{}{}{}$ $\endgroup$ – WillO May 17 '20 at 20:52
  • $\begingroup$ @WillO Very insightful, thank you. $\endgroup$ – Jannik Pitt May 17 '20 at 21:14

If I'm reading this question properly, it boils down to this: if we always need to work with normalized states (which represent a fixed total probability), then how can interference enhance or suppress the probability of seeing a particular state?

Hopefully the following example will clear things up. Suppose that you have an interference experiment where, if you shoot a particle through one path, the final state is $$ | \psi_1\rangle = \frac{1}{\sqrt{2}}| \text{up} \rangle + \frac{1}{\sqrt{2}} | \text{down} \rangle$$ while if you shoot the particle through another path, the final state is $$ | \psi_2\rangle = \frac{1}{\sqrt{2}}| \text{up} \rangle - \frac{1}{\sqrt{2}} | \text{down} \rangle.$$ In both cases, the chance of seeing up or down is 50/50.

Now suppose you send the particle in an equal superposition of the two paths. Assuming that this process does not itself introduce extraneous relative phase shifts, the final state is proportional to $|\psi_1 \rangle + |\psi_2 \rangle$, with some normalization constant that depends on the details. Since $$|\psi_1 \rangle + |\psi_2 \rangle = \sqrt{2} \, |\text{up} \rangle$$ we know the normalization factor is $1/\sqrt{2}$, so the final state is $$\frac{1}{\sqrt{2}} |\psi_1 \rangle + \frac{1}{\sqrt{2}} |\psi_2 \rangle = |\text{up} \rangle.$$ Everything has remained normalized as required. But now there's a 100% chance of seeing up and a 0% down, because of interference.

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    $\begingroup$ What is exactly meant by an equal superposition? If it's referring to the equality of probability of getting either $|\psi_1\rangle$ or $|\psi_2\rangle$ upon measurement of an operator whose non-denerate eigenstates they are, the final state could also be proportional to $|\psi_1\rangle+e^{i\theta}|\psi_2\rangle$ for an arbitrary $\theta$. $\endgroup$ – Dvij D.C. May 17 '20 at 21:05
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    $\begingroup$ @Dvij D.C. Good point, I edited! $\endgroup$ – knzhou May 17 '20 at 21:51
  • $\begingroup$ As an alternative refinement, the state $\vert \psi_1\rangle$ is an eigenstate of $\sigma_x$ so measuring spin along $\hat x$ will give “spin-up along $\hat x$” with 100% probability. $\endgroup$ – ZeroTheHero May 17 '20 at 22:49

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