Real meaning of stationary states

1. May 6, 2010

Rick89

Hi, when we consider an N-particle (assume non-interacting) system, say putting them in a box; why do we always say the states of the system (e.g. when counting them to find the "density of states as function of energy") are just the products of single particle stationary states (i.e. energy eigenstates)(maybe symmetrized or antisymmetrized as appropriate) and not count superpositions of single-particle states as independent states of the system? Thanks

2. May 6, 2010

alxm

The stationary states of noninteracting single-particle wave functions form a complete orthonormal set, so it's a basis, and a fairly convenient one.

3. May 6, 2010

Rick89

I know that, but why do we only count the dimension of the Hilbert space (i.e. the number of elements in the basis) exactly? Why is that what we call the number of states?

4. May 6, 2010

LostConjugate

Each "dimension" in Hilbert space is actually each possible argument for the wave function. It is just converted to vector indices for mathematical notation. (Linear Functional Algebra)

For each argument of the function there is a result given by the function.

Though I am not sure what you mean by calling it the number of states. I thought a new state was a new function, not just a new return value of the same function.

5. May 6, 2010

Rick89

Well, not really. If u consider for example the eigenstates for a particle in a box these are countably infinite, then the state space has a countably infinite dimension, not one for each real value of x. I find talking about wavefunction more confusing, why introduce the position basis? Anyway if u prefer, I am saying that psi1+psi2 (where each is an energy eigenstate) is another state, do u agree? we are not counting those (obviously they are infinite) but why exactly?

6. May 6, 2010

LostConjugate

Hilbert Space remains the same, no matter that the eigenstates are non-continuous. That's the way I understand it anyways.

I agree that the linear combination of two states is another state, yes.

The possible arguments are infinite, not just in position space, also in momentum space. Unless there is a position x that does not exist past a certain point, or a momentum p that is not achievable.

Sorry if I am mis-understanding your questions.

7. May 6, 2010

LostConjugate

To answer your original question, a stationary state is a state that does not change with time. It is not a state which is transitioning in any way. The energy is also not changing in time. Eigenstates of the Hamiltonian are stationary states.

It is similar to a classic stationary state in wave mechanics.

8. May 6, 2010

Rick89

Well, I am not sure what you're talking about either... The state space of the particle in box system is the linear space spanned by the eigenstates, I am not sure I see what you mean when you talk about the argument you give to a wavefunction, how is that related? I am saying an obvious thing actually, just that when we count states we only consider stationary states, I couldn't imagine doing otherwise, but I am trying to understand why is that what we want. Let's give an example: you know when we talk about statistical ensamble, we calculate the prob. distribution of a system (say harmonic oscillator) for being in state n (we mean eigenstate) given thermal equilibrium and given constraints and then say multiply by "number of states" with energy around E=hw(n+1/2). To find, in the ensamble, the number of oscillators with energy around E (I clearly mean between E and E+dE).

9. May 6, 2010

LostConjugate

Oh, are you asking how we know the eigenstates span the space?

10. May 6, 2010

Rick89

Second example: you know what a Fermi sphere is? In solid state, free electron model (just consider gas of non-interacting electrons in a box) we count the number of states available by counting the single-particle eigenstates and we fill up each state with a particle ok? (each k state actually counts for two, because of spin). But why are we counting just the eigenstates? couldn't electrons go into linear superp. of these and be in different states?

11. May 6, 2010

Rick89

No! I know very well that, they basically do it by definition of the state space. Do u see what I mean from the examples? Sorry I have not been very clear...

12. May 6, 2010

LostConjugate

Because the eigenstates are the only measurable states, the only states that interact.

13. May 6, 2010

Rick89

Ok, now u are talking about the same thing...
mmhhh because they are the only one that interact eh? not sure what that might mean...why?

14. May 6, 2010

alxm

I'm not sure if this is what you're asking, but:
Take single-particle eigenstates (orbitals) and form a anti-symmetrized many particle-wavefunction (Slater determinant).
Then, if you only have a single determinant, corresponding to only "filling up" the lowest states, then that's an approximation of the ground-state.
But it is not a complete description of the system. A complete description of the system requires all possible determinants for
all possible excited states (singlets, doublets, etc), (i.e. full-CI).

15. May 6, 2010

LostConjugate

For a zero uncertainty state you get an eigenvalue equation.

http://www.physics.sfsu.edu/~greensit/book.pdf [Broken]

See page 105 - 107 Eigenstates As States of Zero Uncertainty

Last edited by a moderator: May 4, 2017
16. May 6, 2010

Rick89

To Lostconjugate: I know that well...I don't see how it can regard this anyway. E.g. If u measure an observable that doesn't commute with the Hamiltonian the state that u get is not an energy eigenstate. This is not adressing my problem anyway.

17. May 6, 2010

Rick89

To alxm: That's more like it. I don't really understand what you're saying though, can you explain more? And anyway aren't u talking about just using higher energy eigenstates above the ground one? It is still concerning eigenstates anyway...

18. May 6, 2010

LostConjugate

I am not sure what your question is...

19. May 6, 2010

Rick89

Mhh let me see... Suppose we have the single-particle spectrum consisting of |1> , |2> etc... Then when filling up the system with N fermions(so that state only one particle) , why do they go into the eigenstates, can't we put a fermion in the independent state |1> +|3> for example? (I don't mean literally |1>+|3>, I am talking about the antisymmetrized version of this...)
Is this concrete example clearer?

20. May 6, 2010

SpectraCat

First of all, the "product of single particle states solution" (I will call these SD's for Slater determinants for brevity) is only an approximation (possibly a very bad one), if the particles can interact with each-other. Interparticle interactions will mix those SD's, so that the exact eigenstates will be (infinite) weighted sums over the various SD solutions. As axlm said, this infinite sum can be calculated to arbitrary accuracy (provided you have the computing power and time to wait) using configuration interaction.

However, it is important to realize that the actual quantum state approximated in the above treatment is still an eigenstate of the full Hamiltonian, and therefore is a stationary state with a time-independent probability density. So, your question about state-counting approaches still stands. I actually am not completely sure that I know the answer, beyond the trite statement that "state-counting approaches are approximations anyway." However, my best guess would be that, since the eigenstates of the full Hamiltonian are a complete basis set, any arbitrary superpostion can always be expressed in that basis, and the probability density (which is what we are really worried about for state counting), will therefore be a simple weighted sum over all the eigenstates, with the counting weights equal to the square moduli of the expansion coefficients of the superposition (whew!). Thus, all those superposition states are intrinsically accounted for by simply counting the eigenstates.