Vectors in Hilbert space

  • Thread starter Bobhawke
  • Start date
  • #1
144
0
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 [tex] \left( \begin{array}{c} 1 \\ 0 \end{array} \right) [/tex], while spin down can be represented by [tex] \left( \begin{array}{c} 0 \\ 1 \end{array} \right) [/tex]

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

[tex] \hat{H} \left | \psi \right \rangle = E \left | \psi \right \rangle [/tex]

and

[tex] \hat{L_{z}} \left | \psi \right \rangle = m\hbar \left | \psi \right \rangle[/tex]

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

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
different bits!

My question is, how can we satisfy both

[tex] \hat{H} \left | \psi \right \rangle = E \left | \psi \right \rangle [/tex]

and

[tex] \hat{L_{z}} \left | \psi \right \rangle = m\hbar \left | \psi \right \rangle[/tex]

when in one case [tex] \left | \psi \right \rangle [/tex] is a 2 dimensional vector, and in the other it is an infinite dimensional vector.
Hi Bobhawke! :smile:

I think the answer is that an electron say could be

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

and the rank-two L acts on the left bit, while the infinite-rank H acts on the right bits. :smile:
 
  • #3
144
0
Ah of course I should have seen that.

Thanks tiny tim!
 

Related Threads on Vectors in Hilbert space

  • Last Post
Replies
6
Views
2K
Replies
14
Views
1K
Replies
8
Views
787
Replies
4
Views
841
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
5
Views
5K
Replies
3
Views
776
  • Last Post
Replies
18
Views
3K
Top