What is meant by singlet and triplet state of an electron system?
Thanks in Well advance....
The singlet state of a two electron system is the set of possible states where the total spin is 0:
It's called the "singlet" because there is only one state with total spin, 0.
The triplet is the set of states with total spin equal to 1:
Again, it's called the "triplet," because there are three states with total spin equal to 1.
I hope someone can provide me with a really dumbed-down explanation of all this (preferably written in terms of electrons having spin "pointing up" or "pointing down". That's the level I'm coming in at).
I understand that in a singlet state there is only one state, which has total spin = 0. In other words, there's only one singlet state which has either: spin up/spin down or spin down/spin up electrons (is that right? Or is it always either spin up/spin down or spin down/spin up?)
Now, there are three triplet states. In the triplet states, the electrons both have spin that points up or down. (Again, is that right?)
Sorry for asking what is probably a very stupid basic question.
Oh, one other thing. How can:
be equal to both 0 and 1?
The one in the triplet should be up-down + down-up. The minus sign is probably a typo.
I could give it a try =):
I'd have preferred to use the arrows and tex-editor but when previewing it it got all wrong and buggy so I'll go for this notation instead: |+ +> means that both electron spins are up, |+ -> means that first spin is up, second is down.
When combining spin you get the possible outcome for the total S = - |s(1) - s(2)| , ... s(1) + s(2) in integer steps. Thus, for two particles that both have spin 1/2 we only get the possibilities: S = -|1/2 - 1/2|, ... 1/2 + 1/2 = 0 and 1
The total m-value for spin can take on values from -S to S. Thus we get:
for S = 0
m has to be 0, as you said. Now, this can be fulfilled by any kind of linear comb. of |+ -> and |- +> since these have total m = 0. As you said yourself, there is a state where m= 0 for S=1 too, so that state also has got a linear comb. of |+ -> and |- +>.
The comb. for S= 0 is |+ - > - | - +> (omitting the norm. constant)
Now, for S=1 we have got three different m=values:
m= -1 can only be "built" by |- -> since -1/2 -1/2 = -1
m= 0 is here represented by |+ - > + | - +>, notice the sign change. This ensures that this is orthogonal to the other combination where m = 0.
m = 1 can only be "built" by |+ +> since 1/2 +1/2 = 1
I hope that this gives you a grasp of the idea.
Thanks for your reply TorKjellsson, but I'm still a bit lost. I can follow what you've written (I think) but I'm confused to the actual physical meaning of something like "|+ - > - | - +>"
To me, I understand that "m = 1 can only be "built" by |+ +> since 1/2 +1/2 = 1" in that in my mind there are two electrons, both with spin in the same direction, giving a total spin of 1. Similarly, two |- -> spin electrons give you spin of -1.
But I don't understand "|+ - > - | - +>". Are there 4 electrons here? Or does it refer to the initial ground state and the excited state? Or have I got the picture in my head completely wrong? :(
|+ -> means that the first electron has spin up, the second has spin down.
|- +> means that the first electron has spin down, the second has spin up.
A combination of these, such as |+ -> + |- +> means that the two electrons are in some sense pointing both up and down at the same time. Let's say that we measure the spin of either electron in this state, then we get that it points up 50% of the time and points down 50% of the time. On average, we thus get 0.
This is one of the corner stones of QM, a system can be in a mixture of different states before being measured. When measuring the state we have different probabilities to obtain one of the "logical" states, that the electrons are pointing in one (individually) direction.
Now, you might also wonder how |+ -> - |- +> differs from the previous combination when we measure on it. The answer is that it depends on what we are measuring.
If we measure S_z, we get the same answer. m = 0 in both cases.
If we measure S² we get different answers, [tex] 2 \hbar [/tex] and 0. It is not surprising since even though the states have the same total m, they have different total S.
Did it clear some things up?=)
Aren't we all. This is quantum superposition, which is very strange and probably unintuitive. In some way, the system is in both states at once: it is simultaneously in a state where the first electron is up and the second down, and also in a state where the first electron is down and the second up. But this explanation doesn't really tell you much, in particular the important difference between "|+-> + |-+>" which is part of the triplet and "|+-> - |-+>" which is the singlet. You can see that qualitative explanations are hard to provide here: I feel like the best way to get some understanding is to understand the math that describes how these states evolve and respond to measurement.
Thanks for all your help, but I fear that I'm getting increasingly out of my depth!
To be perfectly honest, I'm not even clear on the difference between S and m. S is the total spin, and m is......?
The more I go into this, the less I seem to understand. I'd thought that in the singlet state, that the total spin was 0 because both electrons spin were pointing in opposite directions. Now, from what I can gather, you're saying that in the singlet state the total spin is 0 and both electron spins are "pointing both up and down at the same time". And there's a triplet state where the total spin = 1, but similarly both electrons spin are "pointing both up and down at the same time".
In the singlet and triplet states, when both electrons are both pointing in both directions...are they supposed to be pointing in opposite/identical directions depending on whether it's the singlet or triplet state?
Sorry for leading you on a merry ride here. If you feel like one more stab at getting me to understand this, you're a braver man than me! ;)
My brain hurts.
The best thing is to read some texts on angular momentum, I think, but I'll give it a try here. Although, don't take all my words for granted since I can only present my point of view from elementary/half-advanced courses.
One of the many peculiar things that QM brings is that some quantities are quantisized, as you probably are aware of. This comes from saying that we can describe a system by a PDE which with boundary conditions gives integers and half-integers, called quantum numbers. These numbers quantisize angular momentum and energy.
Now, spin is a type of angular momentum. Picture this as a vector with three components. One can also see that aside from the fact that spin has to be quantisized, also one of the components need to be so. The convention is to take the z-direction. So we have to quantum numbers, let's call them s and m. (sometimes, one calls the m that is connected to spin for m_s, since we have a "m" for any type of angular momentum. for instance, orbital angular momentum is called l and this m can be called m_l)
Now, the condition on m is that it can only take on the values -s,-s+1, ... , s-1, s. All this comes from the PDE that we solve, from the Schrödinger W.E.
Thus, to describe an electron (let's omit orbital ang. mom) we need to consider both spin and m, the z-direction of the spin.
Try to accept that these quantities describes the electron and it might be helpful in the future.
Now, for an electron the spin turns out to be 1/2. From the condition on the z-component of the spin, m, we see that m takes the values -1/2 and +1/2. You can picture this as the z-component of the spinvector pointing up and down.
If you have two electrons in a system, the total spin is computed much like the "m"-value. It goes from -|1/2 - 1/2|, integer steps, 1/2 + 1/2, which in this case is trivially 0 and 1.
Now, the combined m satisfies these two conditions:
1) it is the sum of m of the first particle and m of the second
2) it goes from -(total S) up to (total S) in integer steps. (just as for the ind. electrons)
from here you know the story, S=0 only has m=0 , one state so it is called singlet state. S=1 has got three different m, so it's the triplet state.
I think that you also grasp the idea of |+ -> and |- +>, Just as these are different from one another, so is |+ - > + | - +> and |+ - > - | - +>. in both cases, the first and second electron is pointing both "up" and "down" but the composition is different, right? This makes a measurable difference. This is why it might be helpful to think of the mathematics, think of the two vectors (1,0) and (0,1). Surely you see the difference of (1,0) + (0,1) and (1,0) - (0,1) ? The first is (1,1) and the second (1,-1), completely different things. think of this as plausible in QM and why the two cases are different.
Now, how to actually find that for S=1 it is |+ - > + | - +> and S= 0 |+ - > - | - +> is another story. I mean, it could have been the other way around? Or why not just |+ -> for S=1 and |- +> for S=0?
One way to show this is using ladder operators, but for the moment maybe you can just accept it? =P
It got terribly long but I still hope that some of your questions are answered.
Thanks again, TorKjellsson.
I think I just have to accept that what QM tells us can't be pictured neatly in my head by images of a couple of electrons with pointy arrows on them indicating their spin!
I think I understand as much as I actually need to understand, for now at least :P
When I had this problem one told me to refresh the screen before previewing and it worked.
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