Finding Eigenstates of Spin Operators in Quantum Mechanics

SunGod87
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[SOLVED] Quantum Mechanics - Spin

Homework Statement


Problem is attached.



Homework Equations



The Attempt at a Solution



The first part is seemingly straight forward. Measurements are +/- hbar/2, both with probability (1/sqrt[2])^2 = 1/2 of being observed.

For the next part I have written the operator S_x as:

S_x = 1/2 [S+ + S-]
Where S+ and S- are the raising and lowering operators respectively.
ie. S+ = S_x + i S_y and S- = S_x - i S_y
Then using:

S+ | s,m > = [s(s+1 - m(m+1)]^(1/2) hbar | s,m+1 >
S- | s,m > = [s(s+1 - m(m-1)]^(1/2) hbar | s,m+1 >

With s = 1/2 (this is clear from the first part, since we have the eigenvalues of S_z (m) = -1/2 and 1/2 and -s <= m <= s in integer steps.

I obtain
S+ | 1/2 > = 0
S- | 1/2 > = hbar | -1/2 >
S+ | -1/2 > = hbar | 1/2 >
S- | - 1/2 > = 0

So
S_x | 1/2 > = hbar/2 | -1/2 >
S_x | -1/2 > = hbar/2 | 1/2 >

and
S_x | psi > = hbar/2 1/sqrt[2] ( | 1/2 > + | -1/2 > )
So the measurement is simply hbar/2

For the final part (where I become stuck!)
S_y | 1/2 > = 1/2i (S+ - S-) | 1/2 > = ihbar/2 | -1/2 >
S_y | -1/2 > = 1/2i (S+ - S-) | 1/2 > = -ihbar/2 | 1/2 >

So I am required to find a state vector psi such that:
S_y | psi > = -hbar/2 | psi >
1/2i (S+ - S-) | psi > = -hbar/2 | psi >
(S+ - S-) | psi > = -ihbar | psi >

But is it even possible to construct a state vector out of the spin-up and spin-down eigenvectors to give this result? I can't seem to do it?
 

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SunGod87 said:

Homework Statement


Problem is attached.



Homework Equations



The Attempt at a Solution



The first part is seemingly straight forward. Measurements are +/- hbar/2, both with probability (1/sqrt[2])^2 = 1/2 of being observed.

For the next part I have written the operator S_x as:

S_x = 1/2 [S+ + S-]
Where S+ and S- are the raising and lowering operators respectively.
ie. S+ = S_x + i S_y and S- = S_x - i S_y
Then using:

S+ | s,m > = [s(s+1 - m(m+1)]^(1/2) hbar | s,m+1 >
S- | s,m > = [s(s+1 - m(m-1)]^(1/2) hbar | s,m+1 >

With s = 1/2 (this is clear from the first part, since we have the eigenvalues of S_z (m) = -1/2 and 1/2 and -s <= m <= s in integer steps.

I obtain
S+ | 1/2 > = 0
S- | 1/2 > = hbar | -1/2 >
S+ | -1/2 > = hbar | 1/2 >
S- | - 1/2 > = 0

So
S_x | 1/2 > = hbar/2 | -1/2 >
S_x | -1/2 > = hbar/2 | 1/2 >

and
S_x | psi > = hbar/2 1/sqrt[2] ( | 1/2 > + | -1/2 > )
So the measurement is simply hbar/2

For the final part (where I become stuck!)
S_y | 1/2 > = 1/2i (S+ - S-) | 1/2 > = ihbar/2 | -1/2 >
S_y | -1/2 > = 1/2i (S+ - S-) | 1/2 > = -ihbar/2 | 1/2 >

So I am required to find a state vector psi such that:
S_y | psi > = -hbar/2 | psi >
1/2i (S+ - S-) | psi > = -hbar/2 | psi >
(S+ - S-) | psi > = -ihbar | psi >

But is it even possible to construct a state vector out of the spin-up and spin-down eigenvectors to give this result? I can't seem to do it?

The attachment has been approved yet so I did not see the question but what you did all seems correct (disclaimer: I did not check all the coefficients but it all looks reasonable). For Sy, here's a trick: simply write psi as c_1 |+1/2> + c_2 |-1/2> and just impose that this be an eigenstate of S_y. That's all that is needed!
 
So I should have (on the RHS)
-hbar/2 [c_1 | 1/2 > + c_2 | -1/2 >]
When I'm done, right?
Or should I have:
-hbar/2 [ | 1/2 > + | -1/2 >]

I'm pretty sure it's the first one, right?

Edit: Maybe not, I'm confusing myself with random coefficients multiplied for our cause and normalisation coefficients; aren't I?

In which case I obtain c_1 = i and c_2 = -i

Here is the question: http://img88.imageshack.us/img88/7308/q4rf5.png

Edit2: Solved, c_1 = 1 and c_2 = -i. Just the eigenvector of the Pauli spin matrix sigma_y, duh!
 
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