# How can we know the state of a quantum system?

One of the postulates of QM states that given a system in a state $|\psi\rangle$ and given an observable $A$ whose eigenstates are $|\phi_i\rangle$, then the state of the system can be expressed as a linear combination of them such that

$$|\psi\rangle=\sum_ic_i|\phi_i\rangle$$

and the probability of the eigenvalue $a_i$ associated to the eigenstate $|\phi_i\rangle$ of coming out when $A$ is measured is determined by $|c_i|^2$.

So far so good. My question is how are the $c_i$ coefficients determined. I mean, if one can only get eigenvalues when doing measurements, and on top of that the system is left on an eigenstate right after that, how can one know the state in which the system is before performing the measurement (and, with that, the probability of getting the different eigenvalues)?

• The position eigenstates are different from the momentum eigenstates for example. So once position is measured, the system is in a position eigenstate, and in a superposition of momentum eigenstates. The values of ci for the momentum eigenstates can be calculated, if we know the relationship between position and momentum eigenstates. – Kenshin May 2 '16 at 3:06
• This is also known as quantum tomography (en.wikipedia.org/wiki/Quantum_tomography). As you pointed out, the interesting part is getting the phases, which is sometimes known as phase retrieval. – Martin May 2 '16 at 9:21

Experimental determination of $c_i$ values starts with preparing multiple identical systems, then making measurements. From all the measurements, one determines the probabilities, which are the $|c_i|^2$. The square root of the probabilities will tell you the $c_i$ to within a phase factor of the form $e^{i\beta}$, where $\beta$ is real, and may or may not be determinable.

A simple example of multiple identically prepared systems would be a sample of a radioactive mineral with a single parent nuclide.

• What about variational principle? – user36790 May 2 '16 at 11:44
• @MAFIA36790 Do you mean variational method? That is a way to approximate low energy eigenstates and eigenvalues for hamiltonians which don't give exact solutions. That isn't what OP is asking about. – Bill N May 2 '16 at 12:36
• ooh.... got the point. At least we compute the coefficients of the molecular orbitals using the same method... it's efficient and gives quite a good estimate though. – user36790 May 2 '16 at 13:55

Once a measurement ( observation ) is made on a quantum system the system will be in an eigenstate of that property, so if the energy of an electron is measured the electron will afterwards remain in an energy eigenstate ( until some other measurement or interaction occurs ), but if the angular momentum is measured the electron system will afterwards remain in an angular momentum eigenstate, again until some other quantum interaction occurs.

The mixed state equation $\sum_i c_i \phi_i$ is used for time forecasting of quantum systems: solve the time-dependent Schroedinger Equation, write the solution as a mixed-state equation, and then time-evolve to whatever future time you are interested it , then your statement " the probability of the eigenvalue $a_i$ associated to the eigenstate $|\phi_i\rangle$ of coming out when $A$ is measured is determined by $|c_i|^2$." is valid.

• This problem is usually called "state preparation" – zeldredge May 2 '16 at 2:32
• This really doesn't answer the question. OP asks how to determine the $c_i$ values, which this answer fails to address. – Bill N May 2 '16 at 2:55
• he addresses it fine. You define c_i when you prepare/ measure the system. After all, how exactly do you propose to prepare multiple copies of a system to measure c_i in the first place, as your answer suggests? – anon01 May 2 '16 at 3:08