Say we are doing the Stern-Gerlach experiment. Here's my understanding of what decoherence tells us. The particle starts in a superposition of spin up and spin down, but then gets entangled with the measurement apparatus so that we now have a superposition of "was spin up and measured spin up" and "was spin down and measured spin down".

But here's what I don't understand: why can't we ever get the state "was both spin up and down and was measured as both"? It seems that the particle being in both states is a valid possibility, so why can it not be measured as a possibility?


3 Answers 3


Superposition does not mean the particle is in both states at the same time. The state is the superposition. So the fact that you have particles that start in the same state but then get different results upon measurement indicates the state was a superposition of spin states.

Measurements give you single values (ignoring experimental uncertainties/errors here). It is impossible to measure the particle as being both spin up and spin down when you actually measure the spin. QM is weird, but it is not contradictory.

Going a little farther, any quantum state can be expressed as a superposition in some eigenbasis of an observable. This is precisely what we do when we write out the time-independent part the wave function $\psi(x)$ of energy eigenstates for, say, the particle in a box. We are saying that the state $|\psi\rangle$ is a superposition of position states:

$$|\psi\rangle=\int\psi(x)|x\rangle\,\text dx$$

So really we are always "observing superpositions". But that doesn't mean you can get more than one result upon measurement of an observable.

  • $\begingroup$ Thanks for the response. This makes sense, but let me take it a step deeper (which I added as a comment on the original post). Why are "up" and "down" the single values that we can measure. Why couldn't the "single values" we observe be chosen from the set: up, down, both, neither. $\endgroup$
    – Jeff Bass
    Apr 1, 2021 at 17:04
  • $\begingroup$ @JeffBass I'm not sure what you mean by "both" and "neither". $\endgroup$ Apr 1, 2021 at 17:08
  • $\begingroup$ I suppose I'm looking for something a bit more philosophical about what the particle "is" before we observe it. Of course, this is more about the foundations and interpretations than actual theory. I guess my deeper question is "why are these the eigenstates?" It feels like we could make up a new theory of QM that is almost identical to our world, yet it would also be possible for the particle to be detected at multiple locations on a detector screen (for example). Is it simply experimental fact that has directed us to these particular operators and eigenstates? $\endgroup$
    – Jeff Bass
    Apr 1, 2021 at 17:13
  • $\begingroup$ @JeffBass Yes, experimentally no one has ever measured a particle to be at more than one location simultaneously. As for the rest, you kind of moved goal posts on me here. It might be better to ask a new question rather than add questions to this post that makes my answer now incomplete. $\endgroup$ Apr 1, 2021 at 17:16
  • 1
    $\begingroup$ Got it, sorry about that. I had to verify that the choice of operators was based in brute experimental fact before diving into the philosophy of it. Thank you! $\endgroup$
    – Jeff Bass
    Apr 1, 2021 at 17:17

Say you have an electron which can be spin up or spin down, i.e.

$$ | + \rangle, | - \rangle $$ Say you have some sort of measurement device with three states:

$$ | \text{prepared}\rangle, |\text{measure}+\rangle, |\text{measure}-\rangle. $$

Time evolution of the combined electron/device system goes as follows:

$$ | \text{prepared}\rangle| + \rangle \longrightarrow|\text{measure}+\rangle | + \rangle $$ $$ | \text{prepared}\rangle | - \rangle\longrightarrow|\text{measure}-\rangle | - \rangle $$

This is just what it means to make a measurement.

Because time evolution is linear, the evolution of a general electron state is $$ |\text{prepared}\rangle ( a |+ \rangle + b |+\rangle ) \longrightarrow a |\text{measure}+\rangle | + \rangle + b |\text{measure}-\rangle | - \rangle $$ This is the scenario you describe in your question. However, this is not yet decoherence. This is just the standard measurement process.

So, what is decoherence? In real life, measurements do not take place in a vacuum. They take place in an environment, with, say, air molecules bouncing around all over the place. This means that, however the measurement device functions, it will in general be coupled to the environment so that the state of the environment will depend on the state of the measurement device, i.e. they become entangled. (For instance, depending on however the measurement device functions, maybe air molecules will end up in different places depending on the result of the measurement.)

So, coupling the electron/device system to the environment, the full thing evolves as $$ |\text{env}0 \rangle | \text{prepared} \rangle ( a |+ \rangle + b |+\rangle ) $$

$$\longrightarrow a | \text{env1} \rangle |\text{measure}+\rangle | + \rangle + b | \text{env2} \rangle |\text{measure}-\rangle | - \rangle $$

Where $|\text{env0} \rangle$, $|\text{env1} \rangle$, and $|\text{env2} \rangle$ are just the names I am giving to arbitrary states in the Hilbert space of the surrounding environment, which should be thought of as being, for all practical purposes, random.

Now, the Hilbert space of the environment will in general have a very high dimensionality (lots of air molecules). Any two random vectors chosen from a very high dimensional Hilbert space will in general have a very small overlap. In general, if the Hilbert space of the environment is $N$ dimensional, where $N$ is very big, then the overlap will be approximately of the order $$ \langle \text{env2} | \text{env1} \rangle \approx e^{-N}. $$ which is negligibly small. Because the experimenter, practically speaking, is unable to preform measurements on the environment, we should trace out the environment's Hilbert space from our final state to receive a density matrix. This density matrix can be used to calculate the expectation values of measurements. We can then see that, after the measurement is performed, the density matrix of the electron/device system is "mixed," not "pure," and we cannot detect the quantum superposition of the electron/device system. However, they are still, in reality, entangled. It's just that the entanglement is inaccessible because it depends in some complicated way on the environment.

It's interesting to note that decoherence is quantum entanglement taken to an extreme degree, and is therefore a purely quantum effect. However, ironically, the result of decoherence is to reproduce classical expectation values. So you can say that quantum mechanics uses itself to hide itself.

  • $\begingroup$ Thank you. I think the confusion came down to understanding that the possible measurement outcomes are simply given based on experimental fact rather than derived from something deeper. $\endgroup$
    – Jeff Bass
    Apr 1, 2021 at 17:56

The answer is that nobody knows. It is part of the mystery of quantum mechanics. Therefore it is a very good question to ask.


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