# Measure probability of collapse instead of collapse itself

I was researching quantum mechanics recently. A particle can be in two states at the same time, with a probability of "collapsing" to one or the other. The example I learned was an electron's spin.

The probability can be represented like this: $$\alpha|\uparrow\rangle + \beta|\downarrow\rangle$$ where $$\alpha^2 + \beta^2 = 1$$ and $$\alpha^2$$ represents the probability of collapsing up, and $$\beta^2$$ represents the probability of collapsing down.

Here's my question: is it possible to measure these probabilities, instead of just measuring the collapse after measuring?

I know that measuring superposition is not possible, is this the same question?

In general, no. Given an unknown quantum state $$\vert\psi\rangle = \alpha\vert\uparrow\rangle+\beta\vert\downarrow\rangle$$ (where $$\vert \alpha\vert^2+\vert\beta\vert^2=1$$), there is nothing you can do to deduce $$\alpha,\beta$$. I'll give two proofs of this claim (one is not really a proof but rather an informal and intuitive argument, another is actually a general proof but it's perhaps less intuitive):

• It is obvious how you can't deduce $$\alpha,\beta$$ by measuring its spin in any direction: say you measured the spin in some direction $$\vec{n}$$ and got $$+1/2$$. All you can deduce from this is that $$\langle\uparrow_\vec{n}\vert\psi\rangle\neq 0$$. There is no way to deduce anything else. The same goes for the case where you get $$-1/2$$. In the case of spin $$1/2$$ particle, this is a sufficient argument because any Hermitian operator can be shown to be a spin operator in some direction $$\vec{n}$$. Thus, this argument shows that there is no observable that you can measure to deduce $$\alpha,\beta$$. You can raise the objection that you might perform a unitary transformation on the state before measuring it. However, that doesn't change anything. It would simply be equivalent to measuring the spin operator in some other direction $$\vec{n'}$$ on the original state.
• The intuition of the previous argument ought to work in general. The idea is that no matter what result you get in any measurement that you perform on a state, the only thing you can deduce from that is that the overlap of the outcome with the original state was non-zero. You can't deduce the full original state. However, for the sake of completeness, let's also give general proof: Let's say there is something you can do to a given state that lets you know the $$\alpha,\beta$$. Then you can use these $$\alpha,\beta$$ to prepare a new state in the same state $$\alpha\vert\uparrow\rangle+\beta\vert\downarrow\rangle$$. In other words, in effect, you have found a way to clone an arbitrary unknown quantum state. However, it is known that this can't be done, see the no-cloning theorem. The no-cloning theorem simply follows from the linearity of quantum mechanics. Thus, any hope of deducing $$\alpha,\beta$$ from a given arbitrary state is destroyed by the linearity of quantum mechanics. There are other related issues that also arise if you can deduce an unknown quantum state, e.g., you can send superluminal signals using an EPR state.

However, there are two points to note here:

• If someone gave you a state and they told you that they have prepared the state to be in an eigenstate of a non-degenerate operator $$\hat{O}$$, then you can simply measure the said operator $$\hat{O}$$. If the outcome is an eigenvalue $$o$$ then you can be sure that the premeasurement state was also $$\vert o\rangle$$. In your qubit example, this means you'd get to deduce $$\alpha, \beta$$. Notice that this is only possible because you're already told beforehand that the state has been prepared in a given basis. Thus, you are sure that you are not disturbing the state if you measure it in that basis.

• Finally, if you have infinitely many copies of the same state then you can perform quantum state tomography on it and deduce both $$\alpha$$ and $$\beta$$.

The quantum system interacts with its surrounding during measurement. Thus, quantum state cannot maintain its superposed property. It is called Quantum Decoherence. It means that measurement is always associated with collapse. You can find some more details in contents about Quantum open system.

Also, square of coefficients originally represent probability to measure corresponding state(it's Born rule). Knowing these coefficients without collapse is impossible I think. (We can construct a such state, but this is not measurement)

I don't know it is proper answer for orginal purpose of your question.

• You should add that the amplitudes are determined by repeated experiments/collapse. That's how they are determined. Commented Apr 17, 2022 at 5:10
• Yes. In the N measurement, fluctuation from expectation has O(sqrt(N)) order, so we can determine amplitude approximately. Commented Apr 17, 2022 at 5:14

A particle can be in two states at the same time, with a probability of "collapsing" to one or the other.

No, a particle only ever exists in one state, which can be expressed as a linear combination ("superposition") of some basis states. Superposition is addition (+), not disjunction (or).

When we say that the spin state of an electron is $$\alpha\lvert\uparrow\rangle+\beta\lvert\downarrow\rangle$$, we are not saying that the electron's spin is both up and down at the same time, nor are we saying that it could be either up or down, not knowing which. Instead, we are saying that the electron's spin is given by some other state, which does not have a well-defined value for the spin in that direction.

Also, the probability is given by $$\lvert\beta\rvert^2$$, and the normalisation condition is $$\lvert\alpha\rvert^2+\lvert\beta\rvert^2=1$$; $$\alpha$$ and $$\beta$$ need not be real.

Here's my question: is it possible to measure these probabilities, instead of just measuring the collapse after measuring?

It is the nature of random variables that making a single measurement is never enough to determine the probability distribution—you cannot determine the probability that a coin flip will yield heads by flipping the coin a single time. However, if we prepare an ensemble of systems in the same state, then by making repeated measurements, we can estimate the probabilities.

I know that measuring superposition is not possible, is this the same question?

Depending on what you mean by "measuring superposition," one could argue that it is impossible to not measure a superposition, since any state could be expressed as a superposition of some basis states.