# Quantum state: how many bases?

1. Nov 15, 2008

### fisico30

A single particle quantum state $$\Psi$$ is given a by a wave function (vector) representable as a linear superposition of eigenvectors $$\Upsilon$$ weighted by a number proportional to their probability of occurrence:

$$\Psi$$=a_1 $$\Upsilon_1$$+a_2 $$\Upsilon_2$$+.....

In the case of the observable "spin", there are only two eigenvectors, which constitute a 2 dimensional basis.
In the case of the observable energy or momentum, there are more than 2 eigenvectors (n-dimensional basis)

What determines which basis is used in representing the state $$\Psi$$?
Is it just a matter or choice, based on which observable we want to know?

Also, if $$\Psi$$ is represented in with two different basis sets, then the components of the sets must be derivable from each other....but it seems hard to be able to express all the energy, momentum, position eigenvector in terms of only the two spin eigenvectors...

thanks!

2. Nov 15, 2008

### tim_lou

well, I'm not sure what you mean by number of basis. For a general quantum system, the basis would have to be infinite. When people talk about spin states, they are basically talking about a subspace of the Hilbert space (while neglecting momentum, energy..etc). Of course, the Hilbert space contains much more info than just spin, and the two basis of spin isn't a basis at all. It's only a basis for part of it, or a quotient of the Hilbert space.

As for what basis to use, I think it is merely for convenience due to the postulate of quantum mechanics. It says that for any stationary state, the measured value of an operator is one of those eigenvectors. So, if you are talking about angular momentum, it is more convenient to talk about angular momentum eigenstates. You can of course use another basis that has nothing to do with angular momentum, but then you don't get too much information about how this basis react to the angular momentum operators.

3. Nov 15, 2008

### fisico30

Thanks tim_lou.

let me see if I get it:

we have a wavefunction $$\Psi$$ which belongs to Hilbert space.
Every operator has a set of eigenfunctions (eigenvectors) that all belong to the Hilbert space too, and also form a possible orthonormal basis set (a subset of the Hilbert space).

Different operators have their own set of eigenvectors.

The function $$\Psi$$ can be represented using any of those basis sets.
If we chose the basis set for the spin operator, its basis set is formed by only two eigenfunctions....

4. Nov 15, 2008

### tim_lou

Yeah pretty much. In fact, for any hermitian operator, we can always choose a complete set of orthonormal basis. so if our system has a 3 dimension hilbert space, and I have an operator that has 2 eigenvectors, then I may be able to form, say an orthonormal basis, a,b,c so that a,b has some certain eigenvalue and c another eigenvalue of my operator.

However, I feel my explanation might be slightly confusing. So, let me say this in another way. The dimensions of our Hilbert space depends on the problem in our consideration. If the Hamiltonian only has internal spin terms (with spin 1/2), then the Hilbert only has two vectors as a basis. But really, we are just ignoring the other aspects of a particle (momentum and other things), so we are not probing all the details of the space.

In general, if my hamiltonian contains two terms, lets say H1 + H2, both have nothing to do with one another (they are commuting), then I can write a state as a product (or a linear combination of a product) like |n>⊗ |m>. For example, let H1 be a hamiltonian of one particle and H2 for another. Then I can have states like |1>⊗ |2>, it means particle 1 is in state 1 and particle 2 is in state 2. The ⊗ symbol (the tensor product) is just an abstract way of writing the fact i mentioned above. Where 1,2 are just labels. If the number of states for particle 1 are 5, and the number of states for particle 2 are 20, then I'll have a basis with 5*20=100 vectors.

If now, I don't care about the second particle. I just wish to study the first particle, then I only need to consider states of the form |n>, and in my study, I only need to work with a basis that has 5 vectors instead of 100. That simplify a lot of unnecessary work.

Hope that clarifies a bit.

Last edited: Nov 15, 2008