Understanding Single Particle States in Quantum Mechanics

[broegger]

Hi,

I was just reading up on some QM, and I was wondering this: under what circumstances can you treat a bunch of electrons as occupying "single particle states", and when do you have to use one wavefunction depending on all the coordinates, psi(r1, r2, r3, ...)? Hope you know what I mean...

 

[Perturbation]

One state vector would describe the whole system if all particles were identical. The position eigenstate for the whole system of N particles is then $|\vec{x}\rangle =|\vec{x}_n\rangle\otimes\cdots\otimes |\vec{x}_1\rangle$ (corresponding to the position operator $Q=Q_n\otimes\cdots\otimes Q_1$) and the wave-function $\langle\vec{x}|\psi\rangle =\psi (\vec{x}_n\cdots\vec{x}_1)$, as you said. The intepretation of a state vector/operator is that it describes an ensemble of similarly prepared systems, not necessarilly a single particle.

The Schrödinger equation is still the same for the whole system, $H|\psi (t)\rangle =i\hbar\tfrac{\partial}{\partial t}|\psi (t)\rangle$. The only difference is that in, for example, coordinate representation the Hamiltonian gives the energy of the whole system and becomes for, non-relativistic, non-interacting particles (as Gokul pointed out below [thanks, I didn't realize that for some reason])

$$H=\sum_{i=1}^N\left[\frac{-\hbar^2\nabla_i^2}{2M_i}+V_i(\vec{x})\right]$$

Edit: But yeah, the non-interacting system can be treated individually, but generally one should treat the system as a whole, using a state vector to describe the whole ensemble.

 

[Doc Al]

I think the question is: Under what conditions can a beam of electrons be treated like an esemble of single electrons versus having to treat them as (one instance of) a multi-particle system? As long as the electron-electron interaction can be ignored, you can treat the beam as being a collection of independent particles.

 

[Gokul43201]

The only difference is that in, for example, coordinate representation the Hamiltonian gives the energy of the whole system and becomes

$$H=\sum_{i=1}^N\left[\frac{-\hbar^2\nabla_i^2}{2M_i}+V_i(\vec{x})\right]$$

In general, the Hamiltonian will also have terms like V_i,j - the interaction terms between particles. In the absence of these terms we can solve the single-particle SE to find the single-particle eigenstates and construct the multiparticle state as the product of single particle states. The interaction terms prevent us from being able to do this, and we may have to resort to either a clever approximation (for weak interactions) or a more clever guess of the many-body wavenfunction (for strong interactions).

 

[broegger]

I see. Thanks.

Oh, just one more thing: how do you construct the state vector for the whole system from the individual single particle states? Is it like a tensor product (I'm unsure of the mathematics here.)

 

[broegger]

And yet another thing: even though the electrons doesn't interact, they still have to obey the Pauli exclusion principle, right? So we can't consider them independently after all, or...?

 

[Perturbation]

You would write the multi-particle state-vector as a tensor product of the individual state vectors.