# Dimension of Hilbert space (quantum mechanics)

Haye

## Homework Statement

Consider the states with the quantum numbers n = l = 1 and s = 1/2
Let
J = L + S
What is the dimension of the Hilbert space to describe all states with these
quantum numbers?

## The Attempt at a Solution

I believe the dimension of the Hilbert space is dependent on the number of base vectors.

If I am correct, the 'spin-angle functions' would give me the number of bases vector, and thus the dimension of the Hilbert-space.

This is dependent on the quantum numbers s and ms, l and ml and j and mj.
s=1/2, so ms=-1/2 , +1/2
l=1, so ml = -1, 0, +1
j= 1/2 or 3/2, so mj = -1/2 , 1/2 OR -3/2, -1/2, +1/2, +3/2

For j=1/2 you get 2*3*2=12 different combinations, and for j=3/2 it's 2*3*4=24 different combinations.
This gives 36 different base vectors, and therefore the dimension of Hilbert space would be 36?

I have no idea how to go from here. I have the book "introduction to quantum mechanics" by D.J. Griffiths.

If someone could help me out, it would be greatly appreciated.

Last edited:

Staff Emeritus
Homework Helper

## Homework Statement

Consider the states with the quantum numbers n = l = 1 and s = 1
Let
J = L + S
What is the dimension of the Hilbert space to describe all states with these
quantum numbers?

## The Attempt at a Solution

I believe the dimension of the Hilbert space is dependent on the number of base vectors.
s=1/2 gives two of these base vectors, with m=±1/2.

I have no idea how to go from here. I have the book "introduction to quantum mechanics" by D.J. Griffiths.

If someone could help me out, it would be greatly appreciated.
Is that supposed to be S=1 or S=1/2? The problem statement you gave said S=1, but you seem to be assuming S=1/2.

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Haye
s should be 1/2, sorry for that. Thanks for pointing me in the right direction, I'm a lot closer to the answer now (I think so, atleast).

If I am correct, the 'spin-angle functions' would give me the number of bases vector, and thus the dimension of the Hilbert-space.

This is dependent on the quantum numbers s and ms, l and m and j and mj.
s=1/2, so ms=-1/2 , +1/2
l=1, so ml = -1, 0, +1
j= 1/2 or 3/2, so mj = -1/2 , 1/2 OR -3/2, -1/2, +1/2, +3/2

For j=1/2 you get 2*3*2=12 different combinations, and for j=3/2 it's 2*3*4=24 different combinations.
This gives 36 different base vectors, and therefore the dimension of Hilbert space would be 36?

I am completely unsure if my logic is correct, and I would greatly appreciate it if someone could tell me if I am wrong or not.

vela, thank you so much for pointing me in the right direction at least. I feel a lot less lost than I was.

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