# What is the Triplet state

1. Mar 23, 2009

### Marthius

Mathematically I understand the difference between the singlet state |^v>-|v^> and the triplet state |^v>+|v^> but I can't figure out the physical difference. How are we arranging the electrons to get the 2 different total angular momentums (1 for triplet 0 for singlet)?

A diagram might help if anyone happens to have one.

2. Mar 23, 2009

### alxm

Well, physically, the single-electron wave functions are the same - spatially - regardless of spin coordinate.

So if you have a triplet state, the two electrons cannot be in the same (spatial) state due to the Pauli exclusion principle.

Second, the overall spatial wave function will be anti-symmetric with particle interchange, since the spin component is not. Whereas you have the opposite situation in a singlet state. (Since the total w.f. must be anti-symmetric)

3. Mar 23, 2009

### Marthius

Thanks for the quick reply, but I already understood that much. Where I'm lost is the physical significance of the triplet vs. singlet state in itself, not in regards to the spatial wave function. Basically what are the physical effects of being in one versus the other, but not due to the spatial part of the wave function, only the spin dependent part.

4. Mar 23, 2009

### alxm

Well, in the absence of an magnetic field the answer would be: nothing, mostly. Due to the spatial antisymmetry of the triplet state, you can deduce that it'll be higher in energy than the singlet state. It will also have a larger total angular momentum of course.

But one can actually get a lot, if not most, of the relevant information (to a good approximation) about "closed-shell" systems without ever invoking spin, if one wants to. In heavier atoms jj-coupling comes into play.

5. Mar 23, 2009

### Marthius

Here is where I'm stuck. Imagine an actual He atom with 2 electrons. What is the difference in there arrangement if they are in the singlet vs. triplet state so as to get different total spins.

6. Mar 23, 2009

### alxm

Okay. I'm not sure but I think you're asking something along the lines of how you arrive at the total spin state given a wave function. The answer to that is, you don't. Not in non-relativistic theory. If you deal with the Schrödinger equation, then spin is 'tacked on' as a postulate. It's a coordinate or parameter for your wave function. So you need to decide what your total spin state is before you set about calculating your wave function.

The difference in symmetry means different wave functions, which you get through different boundary conditions. In the triplet case, since this wave function is spatially anti-symmetric, you have the node $$\psi=0$$ at the coordinates $$r_1=r_2$$. In the singlet case you have $$\frac{\partial\psi}{\partial j};\quad j=r_1-r_2$$ at the same points in space.

So it's not a question of 'how the electrons are arranged' as much as 'what arrangement you've assigned to the electrons'. At least if I interpreted your question right this time.

7. Mar 24, 2009

### Redbelly98

Staff Emeritus
Other than having the spins aligned for the triplet state (so S=1, mS=-1,0,+1) and opposed for the singlet (so S=0, mS=0), there isn't any other significance that I'm aware of.

8. Mar 31, 2009

### conway

No, it's not as simple as algined vs opposed.

It's easy to tell the singlet and triplet states apart if you just look at the parallel cases. The problem is the antiparallel cases. There are two of them (check it out on Wikipedia) and we identify one of them with the singlet state and the other with the triplet. Why do we identify the difference term with the singlet and the sum with the triplet state?

9. Mar 31, 2009

### Bob S

The triplet state has a magnetic moment; the singlet does not. See previous thread on the "triplet state question"

10. Mar 31, 2009

### conway

But why does it have magnetic moment if the spins are antiparallel?

11. Apr 1, 2009

### peteratcam

The spins aren't really antiparallel.

Consider first as a reminder, just a single electron, and the eigenstates of the Sx operator.
$$\frac{1}{\sqrt(2)}(|\uparrow\rangle + |\downarrow\rangle)\quad\text{and}\quad \frac{1}{\sqrt(2)}(|\uparrow\rangle - |\downarrow\rangle)$$

Both these states have zero expectation of z-component, because it is a superposition of 'antiparallel' states. But this doesn't mean total spin/magnetic moment is zero. The x-component is well defined, in one case -1/2, the other +1/2.
Total S2 is always $$\frac{3\hbar^2}{4}$$ for a single electron.

Now consider the combined spin of two electrons.
The distinction between singlet and triplet states is the total spin.
Total S2 for triplet is $$2\hbar^2$$. The three states are labelled by their z-projection, being eigenstates of SZ with eigenvalue $$m\hbar$$ where m=+1,0,-1.

In the m=0 case, you know *for sure* that the magnetic moment is not along the z direction, but there is still $$\sqrt{2}\hbar$$ which is smeared (in a quantum way) over the x and y directions.
For m=1 or -1, you know *for sure* that $$\hbar$$ amount of spin is in the z direction, but there is still $$\hbar$$ smeared (in a quantum way) over x and y directions.

For the singlet state, the total spin is zero, and you know *for sure* that all spin components are zero.
Peter

Last edited: Apr 1, 2009
12. Apr 1, 2009

### conway

If this were true then I suppose it would answer my question. But I'm quite sure it isn't true. Looking at the equation the spins of the electrons definitely seem to be antiparallel.

13. Apr 1, 2009

### peteratcam

Don't think of the up and down states as vectors which point up or down - they are not! If you want to think about vectors, you have to take expectation values of the quantities you are interested in.

What is the difference between two vectors pointing the same way, and two vectors pointing the opposite ways?
In one the dot product is postive, in the other, it is negative. Are you happy so far?

So lets evaluate the dot product of the two spins in the two quantum states:
The operator for the dot product (which I will call D) of the two spins is:
$$\hat D = \hat S^1_x\hat S^2_x + \hat S^1_y\hat S^2_y + \hat S^1_z\hat S^2_z$$

So we want to take the quantum expectation values:
$$\frac{1}{\sqrt{2}}(\langle \uparrow \downarrow|+\langle \downarrow \uparrow|)\hat D(| \uparrow \downarrow\rangle+| \downarrow \uparrow\rangle)\frac{1}{\sqrt{2}}$$
and
$$\frac{1}{\sqrt{2}}(\langle \uparrow \downarrow|-\langle \downarrow \uparrow|)\hat D(| \uparrow \downarrow\rangle-| \downarrow \uparrow\rangle)\frac{1}{\sqrt{2}}$$

In the first case, we get $$\frac{\hbar^2}{4}$$;
In the second case, we get $$\frac{-3\hbar^2}{4}$$;
I've missed the working, but if you want to see it just say so.

So in the singlet case, you could say the spins are perfectly antiparallel, but in the m=0 triplet case, they are pointing vaguely in the same direction.

Do not think of the 'spin up' and 'spin down' as describing a vector pointing up and down! Up is not antiparallel to down!

Does that make it clearer at all?
Peter

14. Apr 1, 2009

### conway

Yes, thanks a lot.

I did work through the operator calculations and it comes out the way you said. I also tried to evaluate the total spin energy by using (Sx1 + Sx2)^2 + (Sy1+Sy2)^2 + (Sz1 + Sz2)^2.
(Sorry I'm too lazy to learn Latex.)

I was happy when my calculation checked out. I got your D operator as the cross term. The S-squared terms all added up to 1/4 so I got 6/4. The cross terms are doubled; as you point out, they give -3/4 for the singlet case and +1/4 for the triplet case. So when I double and add them to the S-squared terms, I get 2 for the triplet energy and 0 for the singlet energy.

Thanks again for the helpful pointer.

15. Apr 4, 2009

### conway

I am still trying to work out some calculations to see how this system works and I am stuck on one point. I believe that Sx, Sy, and Sz should work as projection operators that should give me the x, y, and z projects on the spin vector. They seem to work properly when I apply them to an ordinary spinor (a, b). For example Sx applied to (a,b) gives me (ib, -ia) and when I take the dot product with (a*, b*) I seem to get the x component of spin.

I am trying to apply the same operator the the product states, (up*down plus down*up), etc. and I don't get anything that works. I am using

Sx (up*down) = (down*up)
Sy (up*down) = (down*up)
Sz (up*down) = -(up*down)

Any ideas? I think this should somehow work.

16. Apr 5, 2009

### conway

For example the combined state:

(up*up) + (down*down)

should give me the m=0 triplet state along the x or y axis, if I'm seeing this correctly.
Is there an actual operator something like the D operator that Peteratcam showed us several posts back (made up of Sx's and Sy'x for example) that will let me verify this calculation?

17. Apr 5, 2009

### peteratcam

Maybe this quick review will help:
A spin 1/2 degree of freedom has a 2-dimensional hilbert space. All quantum states can be expressed as a linear combination of the two states. The basis for the hilbert space is conventionally taken to be the two eigenstates of the Sz operator, called up and down. Any quantum state can be expressed as (a,b) in this basis, so that 'a' is the coefficient of the up state, and 'b' is the coefficient of the down state. With this representation (1,0) is up, and (0,1) is down. The spin operators act within the twodimensional subspace.

If you consider an extended system, with two spin-1/2 degrees of freedom, the dimension of the Hilbert space is now 4 (that is 2x2, not 2+2). Technically the hilbert space is the tensor product the two single spin hilbert spaces.
In dirac notation, a state of the two spin system where the first spin is in the up state, the second spin in the down state is represented:
$$|\uparrow\rangle\otimes|\downarrow\rangle$$
which is often contracted to
$$|\uparrow\downarrow\rangle$$

For the two spin system, the operator for the x-component of the first spin is
$$S_x\otimes{\mathbf 1}$$
where 1 is the identity operator. The operator for the y-component of the second spin is
$${\mathbf 1}\otimes S_y$$
and the operator for the total S_z component is
$$S_z\otimes{\mathbf 1} + {\mathbf 1}\otimes S_z$$

To save effort, these are normally written:
$$S^1_x$$
and
$$S^2_y$$
and for total S_z
$$S^1_z+S^2_z$$
With this notation, it is understood that the spin operator only acts on the spin it is labelled by.

Returning to your question, the m=0 triplet state which you refer to is an eigen state of total Sz, but it is not an eigenstate of total Sx total Sy, so I don't understand what you mean in the most recent post.

Also, you talk of the spin operators as projection operators, but that is not really correct either. You can take a quantum expectation value of any operator, but this not the same as a projection.

So I'm not fully sure what you are asking, but to understand this stuff it is important to understand the concept of a tensor product of hilbert spaces - if you have MATLAB, there are some nice demos i could suggest.

18. Apr 5, 2009

### conway

Thanks again for the very detailed answer. I'm afraid I'm having trouble clarifying my question for you. I was impressed by what you were able to do using the D operator which you defined several posts ago and I wanted to see if I could handle some basic manipulations. So I tried to write out the operator which would measure the spin in the z direction (and the x and y directions too, if I could.)

I can do it for individual spinors using the S matrices. For any spinor phi, I can evaluate

<phi*| Sz | phi>

and that gives me the z component of the spin direction. It even works for the x and y directions if I use the corresponding operators.

So I wanted to try this with states in what you call the four-dimensional (two-state) system. I think I made a mistake by trying to use a the components of your D operator as my S(z,x,or y) operators. For example I used Sz1Sz2 as my operator and it didn't work. In the sense that I got a non-zero result when I applied the calculation to (up*down) + (down*up).

In your most recent post, you define Sz(total) as the sum Sz1 + Sz2. I see now that this is a different operator from the product of Sz1 and Sz2, which is what I originally tried to use.

So now I've applied your Sz operator to the composite state (up*down) + (down*up) to evaluate the z component of spin. It should come to zero of course. The general method is:

<phi*| operator | phi>

The Sz operator does only two things...the up state is multiplied by 1/2 and the down state by -1/2. Remembering that Sz1 applies only to the first electron and Sz2 only to the second, I multiply through towards the right and I get:

1/2 (up*down) - 1/2(down*up) - 1/2 (up*down) + 1/2 (down*up)

which already comes to zero, so I don't even have to multiply by phi*. So far so good.

If I apply your Sz to the (up*up) case I get 1/2(up*up) + 1/2 (up*up) which is just phi: when I multiply this by phi* I get 1, which is correct for the total up spin.

So this is working out good so far. But now I want to evaluate the x and y components of the spin. I SHOULD be able to do this, because I can do it for individual spinors. By analogy with what you gave me for the Sz operator, I'm going to try and use for the x and y directions:

Sx = Sx1 + Sx2
Sy = Sy1 + Sy2

Let's remember what these operators do: one of them (say Sx) just flips up to down, and the other one (say Sy) flips up to down with a factor of i, and down to up with a factor of -i. And there's always a factor of 1/2 of course...

The particular test case I want to do is (up*up) + (down*down). Because I'm QUITE sure it ought to have a component of spin either in the x of y directions. Let's try the operator multiplications:

Sx1|up*up> = 1/2*|down*up>

Sx2|up*up> = 1/2*|up*down>

Sx1|down*down> = 1/2*|up*down>

Sx2|down*down> = 1/2 |down*up>

When I collect terms and add them up I get |up*down + down*up>. Now I have to multiply by phi*, and right away I can see it's going to zero because the states are orthogonal...right?? in other words <up*up|down*up> = zero, etc. (now I'm starting to wonder...??)

Similar thing happens if I try to evaluate Sy. And yet I'm pretty sure that the state in question (up*up + down*down) should have a net spin either in the x or y directions. Does this make sense?

Hope you can find my mistake.

19. Apr 5, 2009

### conway

Oops. I found my mistake, at least one of them. I'm working on the wrong state. The state I picked has no spin in any direction:

up*up + down*down

is basically the s=1,m=0 state as taken along the x axis.

I think I should get the x or y spin if I take the state

1/2(up*up + i*up*down + i*down*up +/-(down*down))

I'm going to try the Sx and Sy operators on this combination and see if it works.

20. Apr 6, 2009

### peteratcam

First ignore my antepenultimate paragraph in my last post.

The state [(up*up)+(down*down)]/sqrt(2) happens to be the m=0 triplet state along the y-axis.

It is an eigenstate of totalS^2, and also an eigenstate of totalSy with eigenvalue 0.

Remember, if you measure an observable, you will always get an eigenvalue of the corresponding operator as the result of the measurement. The quantum mechanical average obtained by
$$\langle \phi|\hat O |\phi\rangle$$
gives the *average* results of such measurements of O taken in the state phi.

So in the state (up*up+down*down)/sqrt(2), it turns out that the average S_z is 0, the average S_y is 0 and the average S_x 0. (you can check this if you like)
If you examine the average Sy^2, it is 0. But the average Sz^2 = 1, and the average Sx^2=1; This means if you measure Sx, you are equally likely to obtain +1 as -1, similarly with Sz.

(You didn't answer about whether you have Matlab, its just that it is easy to do all these calculations very fast and play around with quantum spins after a few lines of code, which is why i mention it)