How can electrons have the same spin in triplet state?

[Edge5]

Hello, I don't understand how can electrons in triplet state can have the same value of spin? Shouldn't the spins be different because two fermions can't have the same state?
The following picture explain my question in more detail.

 

[ShayanJ]

That's because the spin state is not the only part of the state. Any degree of freedom in a quantum system will have a vector from an appropriate Hilbert state associated to it which describes the state of that degree of freedom. The full state is a tensor product of all the states of all degrees of freedom. For example if we consider that in this case, there are only spin and spatial degrees of freedom, then the full state of two fermions would be $|\Psi(1,2)\rangle=|\Delta(1,2)\rangle_{spin} \otimes |\Phi(1,2)\rangle_{spatial}$. it is the overall state that has to be completely antisymmetric for any finite number of fermions, but each part can be symmetric which means the other part has to be antisymmetric.

 

[PeroK]

Hello, I don't understand how can electrons in triplet state can have the same value of spin? Shouldn't the spins be different because two fermions can't have the same state?
The following picture explain my question in more detail.
View attachment 234567

To add to what was said above. What would happen if you had three electrons and measured the z-spin of each? One could be up, one could be down, but the third one would have to be the same as one of the others. And, atoms can have many more than 2-3 electrons.

So, it's clear that the Pauli exclusion principle applies to the entire state of the electron: spatial + spin.

This manifests itself in the atomic electon shells. In the ground state, you have only one spatial wavefunction. That then allows only two electons in the ground state, as they must have opposite spins. In the next energy level you have four possible spatial wavefunction, so a maximum of eight electrons; and so on.

 

[Edge5]

That's because the spin state is not the only part of the state. Any degree of freedom in a quantum system will have a vector from an appropriate Hilbert state associated to it which describes the state of that degree of freedom. The full state is a tensor product of all the states of all degrees of freedom. For example if we consider that in this case, there are only spin and spatial degrees of freedom, then the full state of two fermions would be $|\Psi(1,2)\rangle=|\Delta(1,2)\rangle_{spin} \otimes |\Phi(1,2)\rangle_{spatial}$. it is the overall state that has to be completely antisymmetric for any finite number of fermions, but each part can be symmetric which means the other part has to be antisymmetric.

Thank you for clarifiying this issue :)

 

[Edge5]

Can you also explain why we don't have a singlet state as I wrote in the question?

 

[PeroK]

Can you also explain why we don't have a singlet state as I wrote in the question?

The state you have is a linear combination of triplet states. In particular, that means the the expected value of the total spin is $1$.

 

[PeterDonis]

Can you also explain why we don't have a singlet state as I wrote in the question?

You have your terminology mixed up. The state you called a "singlet state"

$$
\frac{1}{\sqrt{2}} \left[ ( 1/2, 1/2 ) - ( - 1/2, - 1/2 ) \right]
$$

is actually a triplet state, because its total spin has absolute value $1$, and the middle of the three states you called "triplet states"

$$
\frac{1}{\sqrt{2}} \left[ ( 1/2, - 1/2 ) + ( - 1/2, 1/2 ) \right]
$$

is actually a singlet state, because its total spin has absolute value $0$. In other words, "singlet" and "triplet", respectively, are not the same as "antisymmetric" and "symmetric", respectively.

As for the latter two terms, the state you said was "antisymmetric" (the first one written above) is not. It's symmetric; if you swap the spins in each term (i.e., switch their order, without changing their values), it's still the same state. (If you flip the spins in each term, i.e., change their signs without changing their order, you get minus the original state, but that's not the relevant operation for determining whether a state is symmetric or antisymmetric under particle exchange.) Here is an example of an antisymmetric spin state for two electrons:

$$
\frac{1}{\sqrt{2}} \left[ ( 1/2, - 1/2 ) - ( - 1/2, 1/2 ) \right]
$$

Notice that if you swap the spins in each term, you get minus the original state.