Fourier Transform ,QM and "Particles"

[rpenner]

Suppose you are making a measurement on the system. Do you get the value $$<\psi|H|\psi>$$ as the result of a SINGLE measurement for energy.?

You have a fundamental problem in that you have said each particle is in a pure energy state, then for each particle $$<\psi|H|\psi>$$ is the measurement of a single particle's energy.

If you want to talk about a collection of particles, then you can't add wave functions -- you have to take a tensor product of wave functions. However, because your model is a bunch of non-interacting particles, at every point in time the energy of the ensemble is going to be the sum of the energies of the individual particles which don't change over time. The tensor product of an arbitrary number of pure wave functions is a pure wave function. The tensor product of an arbitrarily large number of energy eigenstates is an ensemble energy eigenstate.

So $$<\psi_{\tiny \textrm{ensemble}}|H_{\tiny \textrm{ensemble}}|\psi_{\tiny \textrm{ensemble}}>$$ is a constant defined by the initial values.

 

[Anamitra Palit]

You have a fundamental problem in that you have said each particle is in a pure energy state, then for each particle $$<\psi|H|\psi>$$ is the measurement of a single particle's energy.

Let's consider the state given by:
$$\psi=c_1\psi_1+c_2\psi_2+c_3\psi_4+c_5\psi_5+.........c_n\psi_n$$ ------------ (1)

$$E_i$$ is the measured value of energy for the state $$\psi_i$$. This state pops up with the probability $$\mid c_i\mid^2$$. We are considering a normalized psi function here,that is, $$\Sigma \mid c_i\mid^2=1$$
For different measurements you are get different values for $$E_i$$ with the probability $$\mid c_i \mid^2$$ even if the system is closed.But then how is the energy changing if the system is a closed one? At the most you may say that for each measurement the system is not a closed for a short period of time when the measurement is being made and there is a scope for the measured value of energy to change due to interaction with the measuring gadget.

How do you interpret the value of $$\langle \psi \mid H \mid \psi \rangle =\Sigma \mid c_i \mid^2 E_i$$?
Do you ever get such a value in a SINGLE measurement?
[With each measurement you may consider a modification of wave function on the same set of eigen functions,only the probabilities may change]

You may examine the issue in the light of what Alphaneumeric has said:

Energy and momentum conservation in quantum mechanics follow from the form of the Hamiltonian and Noether's theorem. As I've repeatedly commented, you haven't even considered the Schrodinger equation, you're messing with plane wave solutions to it without considering the system itself.

No, the energy doesn't change in time. The energy evolution is determined from the Hamiltonian. This follows via the Heisenberg equations, $$\frac{\partial A}{\partial t} = [A,H]$$. If operator A is H then since [H,H]=0 we have $$\partial_{t}H = 0$$ and thus it's expectation value is time independent (just convert to the Schrodinger picture), which is energy. What values the eigenvalues take is irrelevant, since both the eigenvalues and the time evolution 'flow' of the system are determined by the same thing, the Hamiltonian.

You're now showing you don't understand the relevant quantum mechanics, showing I was right to move this to pseudoscience.

 

[rpenner]

Now you are just being sloppy as to what is what.

Are you talking about a pure energy eigenstate or a fixed superposition of pure energy eigenstates?
Are you talking about a single particle or a collection?
If a collection, do the particles interact with each other or do they not?
If a collection, are they distinct or indistinguishable?
If indistinguishable, are they fermions or bosons?

For a many-particle system we have:
$$\psi (x,t)=\Sigma_n\Sigma_j C_je^{-ia(E_jt-p_jx_n)}$$ ---------- (8)
"j" runs over the number of states and "n" over the different particles.

Incorrect.

For " n" particles: we may consider the function:
$$\psi=\Sigma C_j e^{-ia(E_jt-p_jx_j)}dtdx_j$$ ------------ (6)
"j" runs over the particles:j=1,2,3...n

Incorrect, but also contradicts eqn 8 of an earlier post.

You have a fundamental problem in that you have said each particle is in a pure energy state, then for each particle $$<\psi|H|\psi>$$ is the measurement of a single particle's energy.

If you want to talk about a collection of particles, then you can't add wave functions -- you have to take a tensor product of wave functions. However, because your model is a bunch of non-interacting particles, at every point in time the energy of the ensemble is going to be the sum of the energies of the individual particles which don't change over time. The tensor product of an arbitrary number of pure wave functions is a pure wave function. The tensor product of an arbitrarily large number of energy eigenstates is an ensemble energy eigenstate.

So $$<\psi_{\tiny \textrm{ensemble}}|H_{\tiny \textrm{ensemble}}|\psi_{\tiny \textrm{ensemble}}>$$ is a constant defined by the initial values.

Let's consider the state given by:
$$\psi=c_1\psi_1+c_2\psi_2+c_3\psi_4+c_5\psi_5+.........c_n\psi_n$$ ------------ (1)

Isn't this different than the cases you gave before?

 

[Anamitra Palit]

Now you are just being sloppy as to what is what.

Are you talking about a pure energy eigenstate or a fixed superposition of pure energy eigenstates?
Are you talking about a single particle or a collection?
If a collection, do the particles interact with each other or do they not?
If a collection, are they distinct or indistinguishable?
If indistinguishable, are they fermions or bosons?
[\QUOTE]
Let us concentrate(for this post) on a single particle state----a particle in a box. You are performing a measurement on it to find its energy.

Let the eigenfunctions be:
$$(\psi_1,\psi_2,\psi_3,.........\psi_n)$$

We have,

$$\psi=c_1\psi_1+c_2\psi_2+c_3\psi_3+........................c_n\psi_n$$

For a pure eigenstate one of the coefficients has non zero mod= one while the others have zero moduli.
We may also have states where some of the coefficients are non-zero while the others are zero.

Before making the measurement you simply don't know the state of the system. Suppose you make several measurements.
For pure eigenstates you get values(Eigenvalues) like $$E_1,E_2.. $$ etc. For a state where some of the $$c_i{\;} s $$ have zero mod while the the others have non zero moduli, do you get a result like $$\langle\psi \mid H \mid \psi \rangle$$ for a SINGLE measurement?
Please do note that the energy of the system changes for each measurement no matter what answer you devise. You just cannot think of a closed system here.You may of course think of the interaction with the gadget a reason for the system to lose its closed nature for short period of time while the measurement is being made.

 

[rpenner]

Incorrect. Measuring the A of the system forces the system into an A eigenstate. Here measuring the energy forces the system into an energy eigenstate and since H commutes with H, the time evolution of the system does not alter H. And so all future measurements will match the initial measurement.

 

[Anamitra Palit]

Incorrect. Measuring the A of the system forces the system into an A eigenstate. Here measuring the energy forces the system into an energy eigenstate and since H commutes with H, the time evolution of the system does not alter H. And so all future measurements will match the initial measurement.

But with each measurement the wave function will collapse. Your effort seems to be in refuting this idea!

 

[rpenner]

A wave function "collapsing" to the same eigenstate is an idempotent operation.

 

[Anamitra Palit]

A wave function "collapsing" to the same eigenstate is an idempotent operation.

You are not addressing the issue at hand. That's interesting!