Spin 1/2-Raising and Lowering operators question

[tjny699]

Spin 1/2--Raising and Lowering operators question

Hi,

Quick question regarding raising and lowering operators.

Sakurai (on pg 23 of Modern QM), gives the spin 1/2 raising and lowering operators S_{+}=\hbar \left|+\right\rangle \left\langle-\right| and S_{-}=\hbar \left|-\right\rangle \left\langle+\right|.

Acting with the raising operator on, say, the spin down state, you get
S_{+} \left|-\right\rangle = \hbar \left|+\right\rangle. The physical interpretation of this is that the raising operator increases the spin component by one unit of \hbar.

This makes sense to me but when I try to explicitly verify this I run into a misunderstanding.

Let's say I apply S_{+} to \left|-\right\rangle and get \hbar \left|+\right\rangle.
To then "measure" the eigenvalue of this spin-up state, would you not apply the S_{z} operator, which would give another factor of \hbar:
S_{z} S_{+} \left|-\right\rangle = S_{z} \hbar \left|+\right\rangle = \frac{\hbar^{2}}{2} \left|+\right\rangle

Or is it a mistake to apply S_{z} after applying the raising operator? If not, how does the extra factor of \hbar disappear?

Thanks.

 

[Bill_K]

tjny699, I believe most people would say that S₊ is ħ times the raising operator. The raising operator just turns |-> into |+>.

 

[tjny699]

Hi Bill_K,

Thanks for your response.

Do you mean that S_{+} should be defined without the \hbar?

I always see it defined as in the original post. Also, later on the raising operator is defined in terms of the S_{x} and S_{y}, which definitely have to have the \hbar. Or perhaps I misunderstood you?

 

[vanhees71]

That's right, you just have to normalize your states to 1, and everything is fine.

 

[tjny699]

So the correct definition is without \hbar? Is there a reason that Sakurai includes it then?

 

[Bill_K]

S_± and S_z have dimensions of angular momentum and their definition must include ħ. The raising and lowering operators are dimensionless and do not include ħ. If Sakurai calls S₊ the raising operator, I would disagree, they are only proportional to each other.

 

[tjny699]

Thanks, Bill. S_{+} is called the raising operator in a number of sources that I've taken a look at. Naming aside, is it correct to apply S_{z} after applying S_{+} as I did in the first post? Sorry to linger on this but I'm trying to understand how this works in detail.

The spin down state when acted upon by S_{+} becomes the spin up state, and to measure it's spin we apply S_{z}? Does this not give an extra factor of \hbar?

 

[Bill_K]

S₊ is called the raising operator in a number of sources that I've taken a look at.

Then I would have to disagree with every one of them. Unless they are working in units where ħ = 1.

 

[tjny699]

OK, I guess my other question still stands--after applying S_{+} to a spin down state do we need to then apply S_{z} to measure the spin of the resulting state? If so, does this not lead to an extra factor of \hbar

 

[Chopin]

I think I understand your question, but if I've misinterpreted it, let me know.

The normalization of the state doesn't affect the eigenvalue of an operator. To see this, remember the definition of an eigenvalue \lambda:

A\Psi = \lambda\Psi

Now try applying that to a scaled version of \Psi:

A(c\Psi) = c(A\Psi) = c(\lambda\Psi) = \lambda(c\Psi)

So you can see, the eigenvalue of c\Psi is also \lambda, for any value of c.

Applied to your question, if you have an operator S_z and states |+\rangle and |-\rangle, such that S_z|+\rangle = \hbar |+\rangle and S_z|-\rangle = -\hbar|-\rangle, then we have:

S_z(\hbar|+\rangle) = \hbar(\hbar|+\rangle)

So the spin is still just \hbar.