How can we reconcile the different vector dimensions in QM equations?

[Bobhawke]

In QM we require that an operator acting on a state vector gives the corresponding observable multiplied by the vector.

Spin up can be represented by the state vector \left( \begin{array}{c} 1 \\ 0 \end{array} \right), while spin down can be represented by \left( \begin{array}{c} 0 \\ 1 \end{array} \right)

As I understand the Hamiltonian is represented by an infinite dimensional matrix, because there is an infinite number of energy eigenstates. My question is, how can we satisfy both

\hat{H} \left | \psi \right \rangle = E \left | \psi \right \rangle

and

\hat{L_{z}} \left | \psi \right \rangle = m\hbar \left | \psi \right \rangle

when in one case \left | \psi \right \rangle is a 2 dimensional vector, and in the other it is an infinite dimensional vector.

 

[tiny-tim]

different bits!

My question is, how can we satisfy both

\hat{H} \left | \psi \right \rangle = E \left | \psi \right \rangle

and

\hat{L_{z}} \left | \psi \right \rangle = m\hbar \left | \psi \right \rangle

when in one case \left | \psi \right \rangle is a 2 dimensional vector, and in the other it is an infinite dimensional vector.

Hi Bobhawke!

I think the answer is that an electron say could be

\left( \begin{array}{c} 1 \\ 0 \end{array} \right) \sum a_n\psi_n ,

and the rank-two L acts on the left bit, while the infinite-rank H acts on the right bits.

 

[Bobhawke]

Ah of course I should have seen that.

Thanks tiny tim!