I've recently started studying QM and I was told that the quantum state in which a particle is will change as its wave function evolves. So let us say I have a source that emits one photon at a time. If I measure the polarisation of a photon from this source and find it to be along a certain axis(say the y-axis) and if I immediately measure its polarisation,I understand it will still be along the y-axis.

But if I measure its polarisation after some time,then the wave equation will have evolved, giving me a polarisation along another axis. Did I get this part right?

If so, I think this same reasoning can be applied to an electron and its spins (right?) and an electron can switch spin states between measurements. That would mean an electron spontaneously changes its states between up and down as time passes, but this can't be right because we don't really see free electrons constantly giving off photons due to spin flips. So which part of my reasoning is false?

Is it that the changes in properties of a quantum system are caused by measurement? Or did I get it all wrong and nothing about a quantum system changes once a measurement is made?

PS:I am assuming the spin is being measured by a Stern-Gerlach type apparatus.

  • $\begingroup$ How do we know, unless we measure it? The longer the time between measurements then the greater the probability that it will have changed in some way. AFAIK, that's all you can say. $\endgroup$ – user108787 Dec 12 '16 at 13:05
  • $\begingroup$ Hi Praveer, I added in paragraphs for clarity, but if you want to group your questions in a different way, please edit as you think fit. But I do think smaller paragraphs help readability. Best of luck with your question. $\endgroup$ – user108787 Dec 12 '16 at 13:09
  • $\begingroup$ @CountTo10 Right. So I wait for a long enough while and measure again. Will the state change now? I have been told it most likely will(the probability that it will remain in the same state is "negligible"). So what causes this change? Is it the measurement or does the electron do it on its own? Edit: Thanks for the trouble. I think its better this way. $\endgroup$ – pkg Dec 12 '16 at 13:17
  • $\begingroup$ I am absolutely not an expert and I hope you get a proper answer. My understanding is, that it is overwhelming likely than the second measurement will produce a new state. 3 things to consider. 1. The length of time between measurements 2. The longer the time the more chance something, like a magnetic field, will affect the state and 3 a totally random vacuum flucation may affect it. will the electron do it on its own it will go back into superposition of states, but particles are indistinguishable, so the idea of the one particle undergoing a series of events may be misleading. $\endgroup$ – user108787 Dec 12 '16 at 13:37

The quantum state evolves in time, but it only provides probability amplitudes for some observables (a probability amplitude is a complex number whose squared modulus, under proper normalization, gives the actual probability of a specific measurement outcome).

We cannot say that the quantum state describes a physical state as in classical physics, where "things" actually change in time continuously and objectively according to their mathematical description. The QM notion of state is very specific and does not map easily to any such "thing" at all.

Only upon measurement does the value of an observable arise as a classical quantity, real and definite, so it is only when it is measured that one can say something about the actual ("classical") physical state of a quantum system (and only something about what was measured).


The state will change if it does not have a definite energy, that's what the "time dependent Shroedinger equation" tells you. But if it does have a definite energy, it's state is "stationary", meaning that it only changes in total overall phase as time elapses, but a change in overall phase does not change the probabilities of any measurement outcomes. Also, a particular measurement outcome, like polarization, only changes if different polarizations involve different energies. So you need three things to get a change in polarization: time, an indefinite energy for the state of the system as a whole, and an interaction or coupling between polarization and the uncertainty in the energy. Put differently, something about the energy situation has to be trying to make the polarization change, or it won't change no matter how long you wait.

A good way to think of all this mathematically is called a "superposition state," where we say that a state is a superposition (i.e., a weighted sum) of various different states of definite energy (that's the way to have an indefinite energy). Then each of those individual states have a different time dependence, given by the Schrodinger equation. When you superimpose different states with different time dependences, the state as a whole acquires a time dependence that will change the measurement probabilities-- unless what you are measuring has nothing to do with the energies, and keeps its same characteristics for all the different energy states you are superimposing.

What all this means is, you don't have a problem with spontaneous emission of photons, unless the energy is available to make said photons. The way that energy becomes available is if the states that are undergoing transitions involve a change in energy that can account for the photon energy.

  • $\begingroup$ okay. So if I have an unbound electron(not placed in any potential well-finite or infinite), will it change its properties with time? Also, when we say a system(such as a free electron) is in a superposition of different states, do we mean it lies in a sort of state limbo until we measure it? Or is it that it continuously changes from one state to another? $\endgroup$ – pkg Dec 12 '16 at 17:46
  • $\begingroup$ That depends on whether or not it is in a state of definite energy. For example, it could be a state of definite momentum for such an electron. An electron with a definite momentum has a "plane wave" solution, and this does not change with time if there are no fields. Neither will the spin. $\endgroup$ – Ken G Dec 13 '16 at 15:34
  • $\begingroup$ A superposition can indeed be thought of as a state of "limbo" until measured, but it's not necessary to think of a superposition state as fundamentally different from a state with a definite measurement outcome, because often you can have a state that has a definite result for one measurement, that is a superposition state for another measurement (such as can happen with momentum and position in the Heisenberg uncertainty principle). $\endgroup$ – Ken G Dec 13 '16 at 15:35
  • $\begingroup$ So we say that a superposition state is a "pure state", meaning that it is a definite state in regard to something, that something is just not necessarily something we have a way to measure. But there will certainly be a definite measurement outcome if that same measurement has just been done, and if the measurement is an energy measurement and the system is isolated from fields and influences, then the outcomes of all measurements keep their same character for all times after that, including the energy. That means that quantum mechanics respects "conservation of energy." $\endgroup$ – Ken G Dec 13 '16 at 15:36

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