# Explanation of spin 1/2

1. Oct 6, 2008

### Niles

1. The problem statement, all variables and given/known data
Hi all.

I'm reading about spin 1/2, and I am having a hard time understanding it. I hope you guys can fill the gaps. Here is where I stand at the moment:

The basis vectors are "spin up" and "spin down" in the direction of the z-axis. Thus we have for spin 1/2 in the z-direction:

$$S_z = a \uparrow + b \downarrow$$

where S_z is the general state as seen from the z-axis, and up- and down-arrow are the vectors ( (1,0)^T and (0,1)^T respectively).

Up and down arrow have eigenvalues +\hbar /2 and -\hbar /2, respectively, and the probabilities are |a|^2 and |b|^2.

Now we turn to the spin in the x-direction. Finding the eigenvalues and vectors gives us:

$$S_ + ^x = \left( {\begin{array}{*{20}c} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \\ \end{array}} \right)$$

and

$$S_ - ^x = \left( {\begin{array}{*{20}c} {\frac{1}{{\sqrt 2 }}} \\ {-\frac{1}{{\sqrt 2 }}} \\ \end{array}} \right)$$

with eigenvalues \hbar /2 and -\hbar /2, respectively. In my book (Griffiths) it says then that "the generic spinor S_z can be expressed as the following linear combination:

$$S_z = \left( {\frac{{a + b}}{{\sqrt 2 }}} \right)S_ + ^x + \left( {\frac{{a - b}}{{\sqrt 2 }}} \right)S_ - ^x$$

where the probabilities of measing $$S_-^x$$ and $$S_+^x$$ are the constants in front of the vectors.

This I do not understand. Can you tell me why?

****

EDIT: It's section 4.4.1 i Griffith's Intro to QM.

Last edited: Oct 6, 2008
2. Oct 6, 2008

### tiny-tim

spinors

Hi Niles!

$$S_ + ^x = \left( {\begin{array}{*{20}c} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \\ \end{array}} \right)\text{ and } S_ - ^x = \left( {\begin{array}{*{20}c} {\frac{1}{{\sqrt 2 }}} \\ {-\frac{1}{{\sqrt 2 }}} \\ \end{array}} \right)$$​

So $$\uparrow\ =\ S_ + ^z = \left( {\begin{array}{*{20}c} {1} \\ {0} \\ \end{array}} \right)\ =\ \frac{1}{{\sqrt 2 }}\,(S_ + ^x\ +\ S_ - ^x)$$

and $$\downarrow\ =\ S_ - ^z = \left( {\begin{array}{*{20}c} {0} \\ {1} \\ \end{array}} \right)\ =\ \frac{1}{{\sqrt 2 }}\,(S_ + ^x\ -\ S_ - ^x)$$

So $$a \uparrow\ +\ b \downarrow\ =\ \left( {\frac{{a + b}}{{\sqrt 2 }}} \right)S_ + ^x\ +\ \left( {\frac{{a - b}}{{\sqrt 2 }}} \right)S_ - ^x$$

3. Oct 6, 2008

### Niles

Ahh, spinors.. yes.

But the probability of measing either S_x^+ or S_x^- is the square of the absolute value of the constant in front of each term in the linear combination - so for S_x it is the constants to the right of your equal sign (in your post, the last line) and for S_z it is |a|^2 and |b|^2, because we are looking at the whole thing from the z-axis? (I.e. using spin up and spin down of S_z as basis vectors?)

EDIT: And why is it that we wish to write S_z as a linear combination of S_x?

Thanks for taking the time to reply.

Last edited: Oct 6, 2008
4. Oct 6, 2008

### tiny-tim

Same-old same-old …

a2 + b2 = ((a+b)/√2)2 + ((a-b)/√2)2.
To show that you get the same probability no matter what principal axis you use …

some people think there's something special about the z-direction …

but there isn't!

5. Oct 12, 2008

### Niles

The following expression still haunts me:

$$a \uparrow\ +\ b \downarrow\ =\ \left( {\frac{{a + b}}{{\sqrt 2 }}} \right)S_ + ^x\ +\ \left( {\frac{{a - b}}{{\sqrt 2 }}} \right)S_ - ^x$$

1) What is the physical interpretation of this?
2) What is the physical interpretation of eigenspinors in the x- and y direction? Why do we need those when we are defining everything in terms of the z-direction?
3) When I measure S_x, do I find the spin in the x-direction?

Phew, this QM is sometimes a tough nut to crack. Sadly, my book (Griffith's) does a poor job in explaining spin.. he was doing OK up until that chapter.

6. Oct 12, 2008

### tiny-tim

Hi Niles!

(1) The physical interpretation is that if you have a beam of wotsits which go through a z-oriented thingummy with probability a2 of being spin up and probability b2 of being spin down, then if you had put it through an x-oriented one instead, it would have had spin up and spin down with probabilities (a+b)2/2 and (a-b)2/2.

(2) An eigenspinor in the x- direction is a beam which goes through an x-oriented one spin up with probability 1 or 0.

(3) When you measure S_x, you're altering reality!

You find what the spin in the x-direction is after you measure it … but that has no relation (other than an overall probabilistic one) to what it was (either in that or any other direction) before.

7. Oct 12, 2008

### Niles

Great, thanks Tim.

I guess I have to give it some thought for now.. it's not so intuitive, but I guess I have to be careful not to make it harder than it is.

8. Oct 12, 2008

### Niles

Ok, I've thought a little about it.

1) If eigenspinors really are just eigenstates for spin-½ particles, then how do I interpret the eigenspinors in x and y-direction? E.g.:

$$S_ + ^x = \left( {\begin{array}{*{20}c} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \\ \end{array}} \right)\text{ and } S_ - ^x = \left( {\begin{array}{*{20}c} {\frac{1}{{\sqrt 2 }}} \\ {-\frac{1}{{\sqrt 2 }}} \\ \end{array}} \right)$$

How should I understand these two vectors? For example, if I am given a state P for a spin-½ particle, then P is given in the z-direction and it is a 2-dimensional vector in the vector space spanned by spin up and spin down in the z-direction. Where does the above x-vectors come in?

Thanks for being patient. I hope there is an easy way of explaining these things.

9. Oct 12, 2008

### tiny-tim

Hi Niles!

Forget a and b … it's easier to use cos½θ and sin½θ.

Then a beam of wotsits which will, with probablilty 1, go through a thingummy in the cosθk + sinθi direction …

in other words, a beam in eigenstate P = Scosθk + sinθi+, whose spin is always along the cosθk + sinθi direction …

will go through a thingummy in the k direction with probability cos²½θ and in the -k direction with probability sin²½θ,

and we write that as P = cos½θ.Sz+ + sin½θ.Sz-

So Scosθk + sinθi+ = cos½θ.Sz+ + sin½θ.Sz-

Now we can do the same thing for i and -i as we have just done for k and -k … but this time the angle will be 90º - θ instead of θ …

so we must use cos½(90º - θ) and sin½(90º - θ), which are (cos½θ + sin½θ)/√2 and (cos½θ - sin½θ)/√2 …

which is just (a+b)/√2 and (a-b)/√2.

This is the way spinors work …
where a vector would be split into perpendicular components cosθ and sinθ,
a spinor is split into perpendicular components cos½θ and sin½θ.

10. Oct 12, 2008

### Niles

Ok, I think I'm getting it.

You seem to know what you're talking about - what is the exact definition of an eigenspinor other than being eigenvectors to the Pauli spin matrices? I guess we can say they are basis vectors representing the general spin state of the particle, but that is only spinors for the z-direction.

And do they all span the same space?

11. Oct 12, 2008

### tiny-tim

eigenspinors

Hi Niles !

"eigen" doesn't mean "always" (I htink it actually means "proper"), but it might as well …

if a wave function is an eigenvector of the momentum operator, for example, that means the particle will always have a particular momentum.

Similarly, if a wave function (which, remember, is an amplitude times a phase times a spinor) is an eigenspinor, it will always have a particular spin.

If you put it through the right filter, not only will it go through that filter with probability 1, but it will then go through any identical filters, also with probability 1.

Any unit vector can be a basis vector, and any unit spinor can be a basis spinor.

Eigenspinors are no more basis spinors than any other spinor.

An eigenspinor is no more than a wave function which is "purely" a particular spinor.
Any two non-parallel spinors (they don't have to be opposite) will span all the spinors.

For example (they don't normally tell you this, because it's a bit complicated ):

Scosθk + sinθ(cosφi + sinφj)+ = e½iφcos½θ.Sz+ + e-½iφsin½θ.Sz-

12. Oct 12, 2008

### Niles

Re: eigenspinors

1) So the spin up and spin down spinors in the z-direction span the same space as the following spinors?:

$$S_ + ^x = \left( {\begin{array}{*{20}c} {\frac{1}{{\sqrt 2 }}} \\ {\frac{1}{{\sqrt 2 }}} \\ \end{array}} \right)\text{ and } S_ - ^x = \left( {\begin{array}{*{20}c} {\frac{1}{{\sqrt 2 }}} \\ {-\frac{1}{{\sqrt 2 }}} \\ \end{array}} \right)$$

2) This is my last question, then I'll let you off the hook (pun intended, indeed :rofl:).

Let's say we have an eigenstate of the Hamiltonian (normalized!) given by the following:

$$\left| \psi \right\rangle = \frac{1}{{\sqrt 2 }}\left| 1 \right\rangle + \frac{1}{{\sqrt 2 }}\left| 2 \right\rangle$$

We know that $$\left| 1 \right\rangle$$ and $$\left| 2 \right\rangle$$ represent stationary states with different energies (please correct me when I am wrong). The square of the constant in front of each term is the probabilty that a measurement will yeld an energy associated with each stationary state. Does it make sense to write the following?:

$$\widehat H\left| \psi \right\rangle = E\left| \psi \right\rangle$$

And what energy must I write? The one associated with state 1 or with state 2?

Thank you so much so far! By the way, I think "eigen" means "own" in german.

13. Oct 12, 2008

### tiny-tim

Re: eigenspinors

Hi Niles!

(1) Yes.

(2) $$\widehat H\left| \psi \right\rangle = E\left| \psi \right\rangle$$ only works for each eigenstate individually.

14. Oct 12, 2008

### Niles

Great, I really appreciate you took the time to help me. It is a very nice thing you did.

15. Oct 13, 2008

### Niles

Hi Tim!

Ok, now I am perhaps being too much, but I really mean it this time: It is my final question on this topic!

This time it's about using the correct words. Am I correct to say that $$S_ + ^x$$ is the eigenstate, which corresponds to spin up in the x-direction? If I am correct, then I really think I understand it now.

Thanks in advance. I hope you don't think of me as being annoying.

Best regards,
Niles.

16. Oct 13, 2008

### tiny-tim

:rofl: :rofl: :rofl:​
Hi Niles!

Yes … though I'd put it slightly differently … it's the eigenstate which corresponds to a measurement of spin up with probability 1 if we choose to measure spin along the x-direction.

(i think this bit is going to turn into a PF Library entry … )

I'm slightly wishing I hadn't used cos½θ and sin½θ earlier …

it conceals the fact the the important thing is to remember cos½θ.

Just as you always use cosθ for vector components, so you always use cos½θ for spinor components.

(though, technically, that's with θ being measured around the "small circle" joining the two basis spinors, not necessarily a "great circle" … but that's not a problem in practice, because Sz+ and Sz- are always chosen "in-line", so that it is a "great circle" )

So Scosθk + sinθi+, the spinor at angle θ to the z-direction, is at angle 180º - θ to the -z direction, and so its components in those two directions are cos½θ and cos½(180º-θ), which happens to be sin½θ.​

btw, I strongly recommend looking at Chapter 1 (only) of Penrose and Rindler's "Spinors and space-time, Volume I" for the geometry of spinors … particularly section 1.6.

17. Oct 13, 2008

### Niles

Ahh, perfect! Thanks a lot. You've really made my day!

(You'd better lock this thread ).

Have a nice evening.

18. Oct 14, 2008

### Niles

Re: eigenspinors

Actually I would ask if you could elaborate on one of your answers (so technically I am not asking a new question.. ).

It is regarding the following answer:

In an exercise in my book, we are looking at three atoms, and hence the Hilbert space in this situation is spanned by $$\left| 1 \right\rangle$$, $$\left| 2 \right\rangle$$ and $$\left| 3 \right\rangle$$.

If we pick an arbitrary eigenstate as $$\left| m \right\rangle = \sqrt{0.33}\left| 1 \right\rangle + \sqrt{0.33}\left| 2 \right\rangle + \sqrt{0.33}\left| 3 \right\rangle$$, then I have shown that it is an eigenstate of the Hamiltonian, so it has an associated eigenenergy.

So in this case I am able to write $$\widehat H\left| m \right\rangle = E\left| m \right\rangle$$? What is the difference between this case and the previous case?

I hope I'm not annoying you.

19. Oct 14, 2008

### tiny-tim

Hi Niles!

Earlier you said:
but I think you're now suggesting that $$\left| 1 \right\rangle$$, $$\left| 2 \right\rangle$$ and $$\left| 3 \right\rangle$$ have the same energies.

So that's the difference.

Obviously, any state in the subspace spanned by eigenstates with the same energy will be an eigenstate, also with the same energy.

20. Oct 15, 2008

### Niles

Ahh, I see.. although I am not a chemist, it makes good sense that there is the same energy associated with each atom.

Just a formal question: Am I wrong to say that $$\left| 1 \right\rangle$$, $$\left| 2 \right\rangle$$ and $$\left| 3 \right\rangle$$ span the Hilbert space?

Or is it perhaps better to say that it is the space spanned by the eigenkets of our Hamiltonian (because that is what it is, right?).