# Vectors in Hilbert space

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.

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tiny-tim
Homework Helper
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. Ah of course I should have seen that.

Thanks tiny tim!