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Dimension of Hilbert space (quantum mechanics)

  • Thread starter Haye
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  • #1
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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?

Homework Equations




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.
 
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  • #2
vela
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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?

Homework Equations




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.

Review the addition of angular momenta in your book.
 
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  • #3
15
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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.
 
  • #4
vela
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You're over-counting. The Hilbert space is spanned by either basis vectors ##\lvert l, m_l; s, m_s \rangle## or basis vectors ##\lvert j, m_j \rangle##. In other words, you can look at the individual angular momenta or you can look at the combined angular momenta, but not both at the same time. Also, take the state ##\lvert m_l = 1, m_s = +1/2 \rangle##, for instance. Its the same state as ##\lvert m_j = 3/2 \rangle##.

If you count either set up, you'll see you have 6 basis vectors, so the dimension is 6.
 
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  • #5
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Ah, that makes much more sense, J=L+S ofcourse. I understand it know, thank you so much for your help (:
 

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