# Superposition of basis states.

1. Oct 6, 2012

### Beer-monster

1. The problem statement, all variables and given/known data

This is something I should know, but I keep getting mixed up when I try to think about it.

A quantum state can be written as a superposition of basis states such as $\left | n \right \rangle$

So lets say I have a particle in a potential with discrete energy levels (infinite well, harmonic oscillator etc).

If my particle has equal probability of being in state m or state n I would write the wavefunction.

$$\frac{1}{\sqrt{2}}\left | m \right \rangle + \frac{1}{\sqrt{2}}\left | n \right \rangle$$

Correct?

What if I had two electrons one each in state m and state m? How would I write that wavefunction?

My thoughts were that since the wavefunctions represent a probability amplitude, and I am essentially asking for the simultaneous probability of finding a particle in state n AND state n the answer would be of the form.

$$\left | m \right \rangle \left | n \right \rangle$$

But I've been told that's wrong.

Can someone help me clear this up?

2. Oct 7, 2012

### ehild

In Quantum Mechanics, the wavefunctions belong to systems and not to a single particle. If you have only one particle, you can say that it is in state n or in state m. In case of two non-interacting particles, the wavefunctions depend on the coordinates of both particles, and can be expressed as products of the single-particle wavefuntions and their linear combinations. But you can not ask what is the wavefunction of a single particle. The states of the system can be that both particles are in state n, or both are in state m, and also that the first particle is in state n and the second in m, or the first particle is in state m and the second in state n. When the particles are indistinguishable these two wavefunctions combine into a single state |1n>|2m>±|2n>|1m>.

ehild

3. Oct 7, 2012

### Beer-monster

Sorry I was not clear in my post. I am interested in how to write the wavefunction of the system.

In the first case the system consists of one particle in a superposition of states.

The second case the system consists of two particles in two states.

To check that I'm understanding you correctly:

So if (for some reason) we ignore the symmetry constraints of indistinguishable particles. The system wavefunction would be $\left | m \right \rangle \left | n \right \rangle$

4. Oct 7, 2012

### ehild

The system wavefunction is function of the coordinates, spin,... of both particles. In certain cases it can be built up from single-particle wavefunctions. There are a few possible basic wavefunctions built up from the single-particle wavefunctions |m> and |n>. I do not understand what you mean on "The system wavefunction would be |m>|n>".

ehild

5. Oct 7, 2012

### Beer-monster

I may be getting my wavefunctions and state vectors muddled up.

Basically, I'm asking how to write a multi-particle wavefunction (or multiparticle state vector) for a system.

Specifically, if we ignore the symmetry restrains and assume that our system can be described completely by its principle quantum number (i.e. its energy level |m>). How would you write the state vector for two particles in two separate arbitrary states.

I thought that, as the possible single particle states for each individual particle comprise a Hilbert space, the Hilbert system of the whole system would be a product space of these two spaces. The elements of this product space would be the product of the two state vectors.

So if one particles energy basis is |m> and the others is |n> the basis of the system states would be |m>|n> or |mn>.

Is that correct?

6. Oct 7, 2012

ehild

7. Oct 9, 2012

### Beer-monster

It's my fault ehild. I am not explaining myself very well.

I think I got the answer from here

http://hyperphysics.phy-astr.gsu.edu/hbase/pauli.html

In the first box it states that the wavefunction for two electrons in respective states a and b would be given by $\psi_{1}(a)\psi_{2}(b)$.

Which is basically what I wanted to know for states |m> and |n>. The thing I missed, though I think you tried to warn me, was that I had to include some sort of indicator of which electron (1 or 2) was in which state...probably by noting the position coordinate as such.

$$\psi_{ab} = \psi_{a}(x_{1}) \psi_{b}(x_{2})$$

Of course, if I took the 1/2 spin into account I'd need to write

$$\psi_{ab} = \psi_{a}(x_{1}) \psi_{b}(x_{2}) - \psi_{b}(x_{1}) \psi_{a}(x_{2})$$

Does that seem right?

As a final question: Is there ever a circumstance when it would be acceptable to ignore the symmetry/antisymmetry restraints of bosons/fermions and write the wavefunction as a product?

My understanding is that any scale at which you would have to treat the problem using quantum mechanics, is also as scale where the symmetry constraints would matter. Am I wrong?

8. Oct 9, 2012

### ehild

I understand you at the end, and what you wrote is correct.

If the particles are distinguishable, the wavefunction is the product of the individual particles.
The atoms in a crystal are distinguishable, and if I remember well, the vibration states of the crystal is built up from the individual vibrations of the atoms.

Electrons are indistinguishable so you need to take symmetry into account when building up wavefunctions for the system of electrons from one-electron wavefunctions.

ehild

9. Oct 9, 2012

### Beer-monster

Is there a system that should be treated quantum mechanically but where the particles can be considered distinguishable?

10. Oct 9, 2012

### ehild

As far as I know, atoms are considered distinguishable. And they can be treated quantum-mechanically.

ehild