According to standard quantum mechanics, after a measurement, the original state will "collapse" to one of the eigenstate of the observable, and if one measures it immediately after , the result will be the same. Does the second measurement give the same result in experiment? Is there any real experiment demonstrate this that the immediate measurement gives the same result, or this is just a theoretical assumption?

  • $\begingroup$ Wave function collapse may be the single best example of something that was put into the theory because it was needed to fit experiment but for which there is no theoretical justification (or at least no agreed upon justification). For some more impressive demonstrations of what you are asking about see the Quantum Zeno effect $\endgroup$ – By Symmetry May 6 '17 at 12:50
  • $\begingroup$ thanks for your links. i am also curious about quantum zeno effect $\endgroup$ – dr.bian May 7 '17 at 3:03
  • $\begingroup$ Does the Stern-Gerlach experiment answer your question? $\endgroup$ – mmesser314 Aug 21 '17 at 13:02
  • $\begingroup$ i don't know exactly how a "second measurement" is done, since they all destroyed on the screen during the first measurement. but i would like to know how to achieve the second measurement, if this has already be done in experiment. i read more on this afterwards and i find a concept POVM, that can help to explain this: the measurement i describe in the question is just a subgroup of all the measurements, it is called projection measurement. while POVM is a bigger group. i think it is introduced by physicists just to explain the observed fact that "a second measurement won't give " $\endgroup$ – dr.bian Aug 21 '17 at 16:06
  • $\begingroup$ the same outcome as the first measurement". but still i got more questions. one of them is that povm says the state after povm measurement will be in a certain state. i wonder is this only a theory prediction, or it is also verified in experment that after a povm measurement, the state of the object will be in the one that is predicted by povm theory. since when using povm, people typically just care about the measurement results, not the state after the measurement. so i think this may not have been confirmed by experments(i am not expert on this, maybe "second povm measurement" is also $\endgroup$ – dr.bian Aug 21 '17 at 16:14

The depends upon whether the state evolves after the first measurement. Assume the state $$ |\psi\rangle~=~c_1(t)|1\rangle~+~c_2(t)|2\rangle. $$ where the amplitudes are given by a unitary operator. Assume we have the unitary operator $$ U(t)~=~e^{-iE_{11}t/\hbar}|1\rangle\langle 1|~+~e^{-iE_{22}t/\hbar}|2\rangle\langle 2|~+~(1~-~e^{-iE_{12}t/\hbar})|1\rangle\langle 2|~+~(1~-~e^{-iE_{21}t/\hbar})|2\rangle\langle 1|, $$ so the time evolution of the amplitudes are defined according to $c_1(0)$ and $c_2(0)$. Here the energies $E_{11}$ and $E_{22}$ are the eigenvalues and $E_{12}$ and $E_{21}$ define the transition from state $2$ to $1$ and $1$ to $2$ respectively.

Let us now consider the collapse of a wave function $|\psi\rangle~\rightarrow~|1\rangle$. I am ignoring issues of many worlds or decoherence and thinking in a very Copenhagenish manner. The state then evolves in a time $\delta t$. In this time period it many evolve into $$ |\psi(\delta t)\rangle~=~U(\delta t)|1\rangle~\simeq~\left[\left(1~-~\frac{iE_{11}\delta t}{\hbar}\right)|1\rangle\langle 1|~+~\frac{iE_{21}\delta t}{\hbar}|2 \rangle\langle 1|\right]|1\rangle $$ $$ =~\left(1~-~\frac{iE_{11}\delta t}{\hbar}\right)|1\rangle~+~\frac{iE_{21}\delta t}{\hbar}|2 \rangle. $$ There is then a bit of evolution back into the state $|2\rangle$. After time $\delta t$ the probability the state is in $|1\rangle$ is $1~-~({E_{11}\delta t}/{\hbar})^2$, which is nearly unity. So if we make another measurement we are most likely to find it in $|1\rangle$. There is some $\epsilon$ probability $({E_{21}\delta t}/{\hbar})^2$ it is in state $|2\rangle$.

This is the basis of the so called quantum Zeno effect. If a system is measured and reduced to a state then repeated measurements on time intervals much smaller than the periodicity of the system will force it to remain in that state.

  • $\begingroup$ Take a spin-1/2 ensemble as example. And let's say all the spin are now in the x-y plane, rotating around z (a static magnetic field Bz is applied along this axes). To a particular spin in the ensemble, it is in the state given by |ψ⟩ = c1(t)|1⟩ + c2(t)|2⟩ in your answer, where |1⟩ is the eigenstate of $I_x$ with eigenvale +1/2 and so is |2⟩ (with -1/2). $c1(t)=cos(\omega t)$ and $c2(t)=sin(\omega t)$ where $\omega$ is the "angular velocity" of this rotation. $\endgroup$ – dr.bian May 7 '17 at 3:54
  • $\begingroup$ The above is just a typicle NMR experiment. And say if one wants to measure total magnitization along x axes.Put a coil in x and use induction current to measure the magnitization. In real experiment, people will get a result called "FID"---a plot of magnitization vs measure time. It will be a oscliting signal and this is easy to think accroding to the "classical picture": The magnitizaion is rotating in x-y plane so the observed signal in x is a oscilating signal. $\endgroup$ – dr.bian May 7 '17 at 3:55
  • $\begingroup$ But there is a problem when using "quantum picture",the coil can be seen as a "continous measurement" of spin momentum along x, let's say for a time T. So accroding to standard quantum physics, the zeno effect,at the begining of the measurement, the state will "collapse" to one of |1> or |2> (assume |1>),and a plus magnitization is detected (say A). $\endgroup$ – dr.bian May 7 '17 at 3:56
  • $\begingroup$ Then as the measurement is continues (since the coil is always there), accroding to zeno effect,the state will always remain in |1>, and the measured magnitization will always equals A, so there won't be a osclaliting FID, the FTD will appear to be a straight line equals A (from 0 to T)! This cleary contradicts the real experiement $\endgroup$ – dr.bian May 7 '17 at 3:56
  • $\begingroup$ And although in experiemnt one measures an ensemble of spin, rather than a single spin,I think the above contradiction still presented. I want to know what goes wrong in the above considerations. $\endgroup$ – dr.bian May 7 '17 at 3:56

There are a lot of misunderstandings here:

after a measurement,

One must define "measurement" . Meauserement gives a value from the real numbers from the quantum mechanical probability distribution which is confusingly named a "collapse". The probability distribution is the complex conjugate square of the wavefunction which is a solution of the appropriate quantum mechanical equations for the specific boundary conditions. It is not a balloon. To measure the probability distribution one has to sample many times with the same boundary conditions. Each measurement contributes to building up the distribution.

For example in the scattering problem of double slit with single electrons at the time:

A single electron in the first screen is an instance from the probability distribution for the boundary conditions of the problem, a "collapse" in your terminology. The accumulation with the exact conditions of more and more electrons accumulates the probability distribution which can be fitted with the wavefunction of the system.


the original state will "collapse" to one of the eigenstate of the observable,

Each electron track, if one measured its properties,(momentum, polarization) has a unique value from the probability distribution. When it hits the screen a new boundary condition problem is set up:" electron hitting screen and ionizing", so the momentum etc are lost in the second interaction/measurement.

and if one measures it immediately after ,

We measured it with the spot on the screen. We cannot repeat the experiment so

the result will be the same.

is wrong.

The electron gave energy/momentum to ionize atoms on the screen. It cannot be measured again. Only similar electrons can be set up and in this example each "collapses" to a different direction and it is not repeatable.

Is there any real experiment demonstrate this idea?(that the immediate measurement gives the same result) or this is just a theoritcal assumption?

According to my explanation above this does not make sense within the probabilistic nature of quantum mechanics.

  • $\begingroup$ I am curious of the real quantum experiment but now i can only do some thought experiemt... Assume one can prepare an esenmble of eigenstates of one observable, with all the same eigenvalue.And assume one can measure them seprately, then each measuremt of this observable on each state will give the same result. Is there any real experiment demonstrating this idea? $\endgroup$ – dr.bian May 7 '17 at 4:28
  • $\begingroup$ If no such experiment, and as you pointed out, a measurement will affect the object so the second measurement cannot be repeated, then I think there is a possibility that the "eigenstate" idea in quantum mechanics is an illusion, since it never appears in real experiment. $\endgroup$ – dr.bian May 7 '17 at 4:28
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    $\begingroup$ It appears in experiments with many particles in the same boundary conditions, not with one particle.That is why quantum mechanics and its probabilistic nature had to be postulated. Because the quantum mechanical nature is seen in probability distributions, as with spectra of absorption and emission . The spectra distributions could not be explained with classical electrodynamics. $\endgroup$ – anna v May 7 '17 at 4:57
  • $\begingroup$ I want to clarify that my answer is about measuring the behavior of single particles . The answer by Ben Crowel is about a system described quantum mechanically, i.e. with an inherent probabilistic distribution. The mathematics describes repeated measurements of the same state, but it is a collective probabilistic quantum mechanical phenomenon that is the observable. $\endgroup$ – anna v May 9 '17 at 4:01

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