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 [itex]S_{+}=\hbar \left|+\right\rangle \left\langle-\right|[/itex] and [itex]S_{-}=\hbar \left|-\right\rangle \left\langle+\right|[/itex].

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

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

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

Or is it a mistake to apply [itex]S_{z}[/itex] after applying the raising operator? If not, how does the extra factor of [itex]\hbar[/itex] 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 [itex]S_{+}[/itex] should be defined without the [itex]\hbar[/itex]?

I always see it defined as in the original post. Also, later on the raising operator is defined in terms of the [itex]S_{x}[/itex] and [itex]S_{y}[/itex], which definitely have to have the [itex]\hbar[/itex]. 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 [itex]\hbar[/itex]? 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. [itex]S_{+}[/itex] is called the raising operator in a number of sources that I've taken a look at. Naming aside, is it correct to apply [itex]S_{z}[/itex] after applying [itex]S_{+}[/itex] 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 [itex]S_{+}[/itex] becomes the spin up state, and to measure it's spin we apply [itex]S_{z}[/itex]? Does this not give an extra factor of [itex]\hbar[/itex]?

 

[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 [itex]S_{+}[/itex] to a spin down state do we need to then apply [itex]S_{z}[/itex] to measure the spin of the resulting state? If so, does this not lead to an extra factor of [itex]\hbar[/itex]

 

[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 [itex]\lambda[/itex]:

[tex]A\Psi = \lambda\Psi[/tex]

Now try applying that to a scaled version of [itex]\Psi[/itex]:

[tex]A(c\Psi) = c(A\Psi) = c(\lambda\Psi) = \lambda(c\Psi)[/tex]

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

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

[tex]S_z(\hbar|+\rangle) = \hbar(\hbar|+\rangle)[/tex]

So the spin is still just [itex]\hbar[/itex].