# Question About State Collapse and Energy Measurements in Infinite Well

1. Jun 12, 2013

### Jilvin

I am just starting out in self-study for quantum theory, so forgive me if my question seems elementary or completely misguided. In quantum mechanics, every wave function ψ can be decomposed into a linear combination of basis functions in the following manner:

$\Psi = \Sigma{c_{n}\Psi_{n}}$

Now, $\left|c_n\right|^2$ is the probability of finding an energy E_n upon measurement. For a single particle, it seems that conservation of energy would then require E_n upon all subsequent measurements, and therefore should have "collapsed" into a basis state for as long as the particle is in an infinite well. However, I hear that this is true only "immediately after", and that the Schrodinger equation will require the particle to evolve back into a superposition state. However, this does not sit comfortably with me, as it would seem possible now to measure a different energy E_m corresponding to a different basis state, thereby failing to conserve energy. I know that a "weaker" version of energy conservation holds as <H> (the expectation value of the Hamiltonian) remains constant through time, but is it possible, for a single particle, to measure different energy in two successive measurement while it is in the same state? Doesn't this violate energy conservation? Any illumination would help.

2. Jun 12, 2013

### StarsRuler

Well, it depends of the measurement apparatus, a measurement proccess is complex, it is not something for study in the first stages of QM study. There are destructive and non destructive measurements. If the measurement is not destructive, after the measurement, the state doesn´t return to a superposition of states of differents energies: if the measurement process finish and energy is $E_{n}$, then the Schrödinger equation evolution conservates the eigenvalue of the energy.

3. Jun 12, 2013

### Jilvin

Thanks for the response, that was more complicated than I thought! So, if a measurement is classifiable as non-destructive, then it will not return to a superposition of states? This answer is the one that makes intuitive sense to me at my current level of understanding. Because Griffiths, in his discussion of the infinite well (width a) gives the coefficient values as:

c_n = √(2/a)∫sin(nπx/a)ψ(x,0)dx

After a measurement of E_n, you can "claim" that the state is now ψ_n, so that ψ(x,0) for some shifted time parameter is now ψ_n, so that :

c_n = √(2/a)∫sin(nπx/a)sin(nπx/a)dx = 1 (owing to the orthogonality of {sin(mπx/a)})

Therefore, every eigenvalue measurement should yield c_n forever after... there may be a flaw in assuming that you can "consider" ψ(x, 0) to have changed to ψ_n just because your energy measurement is E_n... is this where I am going wrong? I can't seem to reconcile this with the oft-repeated statement that "measurements 'immediately' (whatever that means) after reproduce the measurement with certainty."

4. Jun 12, 2013

### StarsRuler

Well. There are contradictory staments: or wavefunction collapse to an energy eigenfuntion, or it returns to a superposition ( not neccesary the original, of course). The question is that answer can depends of measurement method. An example with the position measurement that maybe can help to understand destructive and non destructive measurements. You can measure the position ( it is a continuous observable, of course, it is not quantized, and not an eigenfunction after measurement, you obtain always an uncertainty in the measurement). If measurement is not destructive, the wavefunction collapses to a wavefunction that is the original wavefunction times a window function that let the values only in the interval of possible position of the wavefunction. If you quickly repeat the measurement, the position is the same. An example of a destructive position measurement: you can fire with a energetic photon ( more energetic, less uncertainty), the electron. If the photon has enough energy, electron can desintegrate. You measured the position, but after the measurement, there is no the same position, ¡¡¡electron doesn´t exist yet !!!. It would be a destructive measurement. But you obtain really a position measurement, the position that it had in the time origin of the measurement. Both of them are really measurements. You can repeat the system in the initial state, let evolve, and obtain a probability distribution, in spite of a destructive measurement. Of course, it depends totally of measurement method. You have to consider the interaction with the measurement apparatus, and there are many interpretation of why happens this non unitary evolution ( the collapse), different of Scrhödinger equation linear evolution. This is the measurement problem. There are many interpretations. Wikipedia does a reasonable summary http://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics. Not all the interpretations can coincide with the explanation I did up. There are interesting debates in this same forums. Discussion about "Schrödinger cat", etc...

5. Jun 13, 2013

### Ravi Mohan

That is nice. I would recommend Feynman Lecture vol III for you.
Correct.
Correct. But it is not the conservation of energy which yields this result. It is because of postulates of Quamtum Mechanics. Measurement is seen as some sort of interaction of some external system with our given system. Thus conservation of energy will mean that the total energy (of course the expectation value) of the external syatem + our system remains same.
Incorrect. You can cite the source if you want. I would like to know in what context it is talking about.
Now there are different measurements corresponding to position, energy etc. The observable corresponding to energy, as we know, is Hamiltonian operator of the system $\hat{H}$.
Now if you are considering the case in which the energy eigen value obtained, during an energy measurement, is the same as the one of the eigen values of the Hamiltonian of square well then the statefunction (or wavefunction) must have collapsed to corresponding eigenstate such that $\hat{H}$$\psi_n(x,t)$ = $E_n$$\psi_n(x,t)$. This is clear.
So let us apply schrodinger's equation and find out what happens
$\iota$$\hbar$$\frac{\partial \Psi(x,t)}{\partial t}$ = $\hat{H}$$\Psi(x,t)$
now after the measurement we have $\Psi(x,t)$ = $\psi_n(x,t)$ corresponding to energy $E_n$. Thus the equation now becomes
$\iota$$\hbar$$\frac{\partial \psi_n(x,t)}{\partial t}$ = $\hat{H}$$\psi_n(x,t)$ = $E_n$$\psi_n(x,t)$

so
$\iota$$\hbar$$\frac{\partial \psi_n(x,t)}{\partial t}$ = $E_n$$\psi_n(x,t)$

on solving
$\psi_n(x,t)$ = $\exp{\frac{\iota E_n t}{\hbar}}$$\psi_n(x,0)$
t=0 is just after the measurement.

Here you can see that as the time goes, the statefunction remains essentially the same (with varying phase factor). Thus the energy is constant with respect to time. The system now is not evolving as superposition of energy eigen states.

We can conclude that if the system is in one of the eigen states of the hamiltonian of the syatem then it remains there for infinite time. This is nothing but uncertainity principle in which $\Delta E$$\Delta t$ > some constant.
Now in this eigen sate $\Delta E$ = $0$ thus $\Delta t = \infty$. Thus these energy eigen states are also known as stationary states.

In time independent schrodinger equation you essentially evaluate these stationary states! You know how stationary state evolves (just a phase factor) so you eliminate the time alltogether. (forgive me if you already know this stuff, but I really get excited sometimes when I explain something

But if suppose you would have measured position of the particle. It would have collapsed to a dirac delta function. Now if you let the systm evolve in time, then it will evolve in superposition of eigenstates of position operator. You can ckeck it yourself from schrodinger equation.
I believe this should sit comfortably now.

I think now you can answer this too.

Last edited: Jun 14, 2013