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Linear combination of states

  • Thread starter positron98
  • Start date

positron98

The following linear combination of states is considered in almost all quantum mechanics textbooks when they try to explain the addition of spin 1/2 and orbital angular momentum. The thing I don't understand is how the left hand side is equal to the right. Please, if you can, explain how.

|j,m + 1/2> = a|lm,+> + b|lm,-> where |+> is spin up and |-> is spin down.


Thanks,

Sam
 

Tom Mattson

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Write the spin and orbital states separately, and the total as a direct product of the two, as follows:

|J,MJ>=|L,ML>*|1/2,1/2>

Then use the ladder operators to work your way down.

Example:

L=1, S=1/2

|3/2,3/2>=|1,1>*|1/2,1/2>

Apply the ladder operator J-=L-+S-, noting that on the RHS the operator only acts on its respective ket.

J-|3/2,3/2>=(L-|1,1>)*|1/2,1/2>+|1,1>*(S-|1/2,1/2>)

and keep working down.

Try to work out the combination for |J=3/2,MJ=1/2>. If you get stuck, post what you came up with and I'll help you through it.
 

Tom Mattson

Staff Emeritus
Science Advisor
Gold Member
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What the hell, I'm feeling ambitious!

J-|3/2,3/2>=(L-|1,1>)*|1/2,1/2>+|1,1>*(S-|1/2,1/2>)

RHS:
J-|3/2,3/2>=31/2(hbar)|3/2,1,2>

LHS:
(L-|1,1>)*|1/2,1/2>+|1,1>*(S-|1/2,1/2>)=
21/2(hbar)|1,0>*|1/2,1/2>+(hbar)|1,1>*|1/2,-1/2>

Putting them together and solving for |3/2,1/2> yields:

|3/2,1/2>=(2/3)1/2|1,0>*|1/2,1/2>+(1/3)1/2|1,1>*|1/2,-1/2>

Try the rest, and let me know if you need help.

Remember: Once you get to J=1/2, you will have a parity change. See me if you need help on that, too.
 

positron98

Thanks Tom for your time and help.

Sam
 

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