# Is “quantum superposition” just a fancy way of saying that a system is in one state or another with some probabilities? [duplicate]

When we say that an electron is in a quantum state that is a linear combination of two eigenstates, one with the probability of 75% and the other 25%, what is actually happening?

1. Is the electron "really" somehow magically in these two eigenstates at once?

OR

1. The electron is in only one of these eigenstates at a time point, we just can't tell which one it is without observing it. And when we observe the electron many times, we find it in one eigenstate in 75% of the time, and in the other eigenstate in 25% of the time, because the electron switches between these two states and just spends three times more time in one eigenstate than the other?

Obviously, I think the second case is correct.

But which case is correct? The 1st or the 2nd?

If the 1st case is correct: That's insane, thank you!

If the 2nd case is correct: Then, isn't the principle of superposition just an "assumption" that should not be taken literally? That is, the electron is not at two eigenstates at once, but one of them at a certain time point, and we just have no way of finding out which state, before observing it? So we just say that it is in an eigenstate X with a probability P(X), such that the sum of probabilities for all eigenstates equal to 1?

I could also, for example, say: I don't know where my friend Max currently is. He could be at school with a probability of 75% and at a bar with a probability of 25%. We all know that Max is not at school and the bar simultaneously, and we can just call him to find out where he is. But before calling him, for the sake of being able to conduct our calculations, we can "assume" that he is at school and the bar simultaneously, with the assigned probabilities, but we actually know that we just made this assumption to be able to carry on with our calculations about Max.

Isn't it just a fancy way of saying that something is in state X with P(X) and state Y with P(Y), and it is in only but only one of these states at all times, however, we just can't find out which state before observing that something, but can merely say with which probability we're likely to find it at a certain state?

Then why is everyone so surprised about the superposition principle?

## marked as duplicate by Emilio Pisanty, Jon Custer, GiorgioP, Cosmas Zachos, Kyle KanosMay 26 at 16:40

It is the first case (using a somewhat loose definition of "being in two eigenstates at once").

The crucial point to understand is that there is a measurable difference between the two cases. In fact, the two cases you mention correspond to two very different kinds of states.

Let us consider for the sake of the argument a single qubit, that is, a state that can be in one (or both?) of two states, that we denote with $$\lvert\uparrow\rangle$$ and $$\lvert\downarrow\rangle$$.

You can have a quantum state such that "The electron is in only one of these eigenstates at a time point, we just can't tell which one it is without observing it". This is what we call a mixed state, and is described by a density matrix $$\rho$$ of the form (assuming equal probabilities for the two outcomes) $$\rho=\frac{1}{2}(\lvert\uparrow\rangle\langle\uparrow\rvert+\lvert\downarrow\rangle\langle\downarrow\rvert).$$ Such a state does not describe a superposition of the two states, rather describes the situation in which you know you have one of those states, you are just not sure which one it is.

This is dramatically different then a state of the form $$\lvert\psi\rangle=\frac{1}{\sqrt2}(\lvert\uparrow\rangle+\lvert\downarrow\rangle),$$ which corresponds to the density matrix $$\rho_\psi=\frac{1}{2}(\lvert\uparrow\rangle\langle\uparrow\rvert+\lvert\downarrow\rangle\langle\downarrow\rvert+\lvert\uparrow\rangle\langle\downarrow\rvert+\lvert\downarrow\rangle\langle\uparrow\rvert).$$

However, crucially, not all kinds of measurements can tell the difference between these two states. Indeed, measuring in the computational basis, which amounts to measuring the probability of getting one of the two outcomes, will give the same exact result in both cases.

But, if you measure in a different basis, the two states will give very different results. Equivalently, you get very different results if you try to evolve the state through some unitary evolution, and then measure the evolved state in the computational basis. On this topic, you can have a look at this other answer of mine, in which you can see an example of a simple calculation on this regard.

The take-home message is therefore that the two cases you mention correspond to very different physical realities. You have to say that the quantum state is somehow "in both states at once", because the results of the measurements are not compatible with the state being in any of the two states, nor with the state being in a "classical mixture" of the two states. More generally, through Bell's inequalities one can prove that no local hidden variable theory can reproduce some results arising from quantum mechanics.

I have just found the answer to my question and if anyone else is wondering what it is (quoting from the Bell's theorem page on Wikipedia):

"No physical theory of local hidden variables can ever reproduce all of the predictions of quantum mechanics."

That is, in my question, I was implying that the randomness associated with the behavior of quantum particles exist just because we don't know enough about them and there must be some parameters that we don't know about that give rise to these seemingly random phenomena. However, John Stewart Bell has showed with his theorem that local hidden variables cannot give rise to these random quantum phenomena.

For more details read the "Bell's theorem" article on Wikipedia.

Thank you very much for the replies so far!

• The answer to my question is not "thank you", but "Bell's theorem", and that's indeed what I'm pointing out in the answer I've added. – Planckturing Dec 25 '17 at 3:16
• hey would be great if you could flesh out this answer a bit – pentane Dec 25 '17 at 3:25
• @pentane Anixx's answer below explains it all very well – Planckturing Dec 30 '17 at 20:35

What none of the replies so far has addressed is the fact that there is a phase involved, not just a probability, and this phase has observable consequences. The state described by the OP could be

$$(3/4)^{1/2}|X\rangle+(1/4)^{1/2}|Y\rangle,$$

or it could be

$$(3/4)^{1/2}|X\rangle-i(1/4)^{1/2}|Y\rangle,$$

or anything else where the amplitudes have squared magnitudes in the ratio of 3 to 1. These are all different states, and they can be distinguished by observing interference.

So the purely probabilistic interpretation #2 given by the OP is definitely wrong, because it basically gives a fuzzy-logic interpretation of quantum mechanics, without the existence of phase or interference.

Interpretation #1 was:

Is the electron "really" somehow magically in these two eigenstates at once?

This is fine, with the caveat that "really" doesn't necessarily mean anything -- and of course the OP was sophisticated enough to realize this, hence the scare quotes.

Keep in mind also that there are not some states that are eigenstates and some that are not eigenstates. Every state is an eigenstate of something. For a particle in a box, we can have an eigenstate of the hamiltonian, which is a standing wave, or an eigenstate of momentum, which is a traveling wave. The standing wave is a superposition of traveling waves, and the traveling wave is a superposition of standing waves.

This is for clearing some missconceptions:

When we say that an electron is in a quantum state that is a linear combination of two eigenstates, one with the probability of 75% and the other 25%, what is actually happening?

What we are saying is:" we can model mathematically the quantum probabilistic behavior of electrons seen in probability distributions from many electrons in the same boundary conditions, by ascribing two quntum states , i.e. two wavefunctions. Th Ψ*Ψ of these superposed wavefunctions reproduce the probabilities seen in experiment."

The electron is in only one of these eigenstates at a time point, we just can't tell which one it is without observing it.

Even so, if we observe it we cannot say from which one of the two superposed wavefunctions the electron came. If it is eigenfunctions of conserved quantities, i.e. solutions of the hydrogen atom for example, a single electron cannot be in two eigenfunctions because it is occupying a specific energy level and energy conservation would be violated in this case. It will be giving its signature by the energy. I am saying that it is better to use the term "wavefunction" for a general discussion of superposition, because there are situations where eigentstates cannot be superposed due to conservation laws.

And when we observe the electron many times, we find it in one eigenstate in 75% of the time, and in the other eigenstate in 25% of the time, because the electron switches between these two states and just spends three times more time in one eigenstate than the other?

As I said above, it cannot be generally eigenstates , just wavefunctions which are superposed, because of conservation laws. (If the operator acting on Ψ does not involve conserved quantum numbers they can be.)

In addition one cannot measure the same electron many times. Once it has been measured, it has interacted and from then on it is described by a different wavefunction. One has to accumulate many measurements with the same boundary conditions to get the probability distribution for the system under consideration.

Is the electron "really" somehow magically in these two eigenstates at once?

No. Each particle is in exactly one state. "Superposition" means that the one state it is in is can be decomposed into multiple eigenstates. If you have a point whose cartesian coordinates are (2,3), that doesn't mean it's magically at the two points (2,0) and (0,3). It means it's at a single point that can be decomposed into (2,0) and (0,3).

The electron is in only one of these eigenstates at a time point, we just can't tell which one it is without observing it.

There are some interpretations in which it is "secretly" in one eigenstate, but those interpretations are not generally accepted, and they require violating properties that many people believe quantum mechanics should follow.

The answer depends on which interpretation of quantum mechanics you are subscribed to. If it's Copenhagen interpretation, for instance, then the electron is magically in two eigenstates at once. If it is Brogle-Bohm interpretation, then it is in one state, but we do not know and cannot know which one (and that depends on unknown initial conditions of the Universe).

The Bohm's interpretation is non-local, so it does not contradict the principle you cited in your answer. It is also completely deterministic.

When we say that an electron is in a quantum state that is a linear combination of two eigenstates, one with the probability of 75% and the other 25%, ....

... what is meant by this? A free electron has an electric field, this field seems to be evenly radially distributed and by this not interesting for changing eigenstates. Furthermore a free electron has a magnetic dipole moment and this moment has an axis from the electrons magnetic north to south pole. And indeed there are experimental setups where the electrons are polarized, means the directions of their magnetic dipole moments are not fully randomly distributed more.

Eigenstates are the expression for the distribution of some values into groups. In our case for example the experimental setup allows to get 75% of the electrons with north pole down and 25% with south pole down. Eigenstates are interesting in the case, that we couldn’t manipulate it’s value by 100%. An example: Placing electrons in an external magnetic field the electrons magnetic dipole moments get aligned all in the same direction. Our result is unambiguous and there is only one eigenstate. In another experiment maybe you

• can observe a 75/25-distribution of the magnetic dipole distribution
• but can’t control the result for each individual electron.

Now about how you measure your outcome. The measure a polarization, be this from an electron or a photon, you use a detector which allows you to measure only the states “passes through” or “get rejected”. Particles go through only with 0 by 90° and 180 by 270° orientation. And what you get is a statistical distribution. The value for each particles has a uncertainty and you never will know the exact value.

The electron is in only one of these eigenstates at a time point, we just can't tell which one it is without observing it. And when we observe the electron many times, we find it in one eigenstate in 75% of the time, and in the other eigenstate in 25% of the time, because the electron switches between these two states and just spends three times more time in one eigenstate than the other?

Let me not agree. It’s an exhilarating feeling to talk about superpositions or change in eigenstates and so on, but the reality is much more trivial. All values of single and uninfluenced (free) particles exist without our observation unambiguous. Our observation influences the particle and there is no way to get the full description of all parameters of the particle without changing some of the characteristics.

The Copenhagen interpretation talks about overlapping eigenstates of particles which collapse under measurement. With the same right you can say that each particle has its values and under a measurement this value gets influenced in a way, that we can conclude only be the help of statistical methods about the measured values.

Now you can decide which story is the easier to understand and which can be told again and again an amazed audience.