# Spin wave function of a system of 2 electrons

1. Feb 25, 2016

### Happiness

Let $\alpha(n)$ and $\beta(n)$ be the eigenfunctions of $S_z$ that correspond to "spin up" and "spin down" for electron $n$ respectively.

(a) Suppose we prepare electron $1$ to have its spin aligned along the $x$ axis. Is its spin wave function
$\chi=\frac{1}{\sqrt{2}}\alpha(1)+\frac{1}{\sqrt{2}}\beta(1)$ or $\chi=\frac{1}{\sqrt{2}}\alpha(1)-\frac{1}{\sqrt{2}}\beta(1)$?

(b) Next, suppose we prepare electron $1$ in the "spin up" state and electron $2$ in the "spin down" state and then mix them together. Is the spin wave function of the combined system of electrons $1$ and $2$
$\chi=\alpha(1)\beta(2)$ or $\chi=\frac{1}{\sqrt{2}}\alpha(1)\beta(2)+\frac{1}{\sqrt{2}}\beta(1)\alpha(2)$ or $\chi=\frac{1}{\sqrt{2}}\alpha(1)\beta(2)-\frac{1}{\sqrt{2}}\beta(1)\alpha(2)$?

(c) Next, suppose we prepare both electrons $1$ and $2$ to have their spin aligned along the $x$ axis and then mix them together. What is the spin wave function of the combined system?

(d) What about the case where electron $1$ has its spin aligned along the $x$ axis and electron $2$, along the negative $x$ axis?

Last edited: Feb 25, 2016
2. Feb 25, 2016

### Staff: Mentor

Conventionally, the one with a + sign.

$\chi=\alpha(1)\beta(2)$. The others are entangled states.

$\chi=\left[ \frac{1}{\sqrt{2}}\alpha(1)+\frac{1}{\sqrt{2}}\beta(1) \right] \left[ \frac{1}{\sqrt{2}}\alpha(2)+\frac{1}{\sqrt{2}}\beta(2) \right]$

Simply rotate the state of electron 2 above by $\pi$ around the z axis.

3. Feb 25, 2016

### blue_leaf77

Since, the OP is talking about electrons and does not seem to consider the spatial wavefunction, shouldn't we take the symmetrization postulate into account, i.e. the wavefunction of fermions should be antisymmetrized?

4. Feb 25, 2016

### Staff: Mentor

Yes, I was negligent.

5. Feb 25, 2016

### Happiness

Does that mean the answer to (a) should instead be $\chi=\frac{1}{\sqrt{2}}\alpha(1)-\frac{1}{\sqrt{2}}\beta(1)$?

Then what is the physical interpretation of the other possibility $\chi=\frac{1}{\sqrt{2}}\alpha(1)+\frac{1}{\sqrt{2}}\beta(1)$?

6. Feb 25, 2016

### Happiness

Since the spin wave function is independent of the azimuthal angle $\phi$, am I right to say that this rotation only affect the spatial wave function and the spin wave function remains the same as the one in (c) as follows?

$\chi=\left[ \frac{1}{\sqrt{2}}\alpha(1)+\frac{1}{\sqrt{2}}\beta(1) \right] \left[ \frac{1}{\sqrt{2}}\alpha(2)+\frac{1}{\sqrt{2}}\beta(2) \right]$

7. Feb 25, 2016

### Staff: Mentor

No, it only affects my answer for the two-electron wave function.

The choice of sign is a question of convention. Using $| +z \rangle$ and $| -z \rangle$ instead of $\alpha$ and $\beta$, setting
$$| \pm x \rangle = \frac{1}{\sqrt{2}} \left[ | +z \rangle \pm | -z \rangle \right] \\ | \pm y \rangle = \frac{1}{\sqrt{2}} \left[ | +z \rangle \pm i | -z \rangle \right]$$
allows to define a right-handed system.

8. Feb 25, 2016

### Happiness

In what way is it affected? Is the answer still $\chi=\alpha(1)\beta(2)$?

9. Feb 25, 2016

### blue_leaf77

For two identical fermions, the wavefunction must satisfy $\psi(1,2) = -\psi(2,1)$, that is, when you interchange the positions of the particles, the wavefunction must change sign. In this sense, a state of the form $\chi=\alpha(1)\beta(2)$ does not satisfy the aforementioned requirement, instead it's satisfied by $\chi=\frac{1}{\sqrt{2}}\alpha(1)\beta(2)-\frac{1}{\sqrt{2}}\beta(1)\alpha(2)$.
No, the rotation operator $\exp(-i \frac{\mathbf{J}\cdot \mathbf{n}}{\hbar} \phi)$ can also act on spin state if $\mathbf{J} = \mathbf{S}$ ($\mathbf{J}$ is a general angular momentum vector operator)..

10. Feb 25, 2016

### Happiness

Looks like for (c)

$\chi=\left[ \frac{1}{\sqrt{2}}\alpha(1)+\frac{1}{\sqrt{2}}\beta(1) \right] \left[ \frac{1}{\sqrt{2}}\alpha(2)+\frac{1}{\sqrt{2}}\beta(2) \right]$

also does not change sign when I interchange $1$ and $2$. Is there anything wrong?

11. Feb 25, 2016

### blue_leaf77

Just take the previous antisymmetrized state for the z direction $\chi=\frac{1}{\sqrt{2}}\alpha(1)\beta(2)-\frac{1}{\sqrt{2}}\beta(1)\alpha(2)$ but now redefine the $\alpha$ and $\beta$ to be the up and down states in $x$ direction.

12. Feb 25, 2016

### Happiness

That would be the answer to (d), right? Because electron $1$ is spin up and electron $2$ is spin down in the $x$ direction.

But for (c), both electrons are spin up in the $x$ direction.

13. Feb 25, 2016

### blue_leaf77

Unless other degrees of freedom are considered, such a situation is not allowed by the symmetrization postulate.

14. Feb 25, 2016

### Happiness

Does that mean we cannot put those two electrons together? And so a wave function of the combined system does not exist?

15. Feb 25, 2016

### blue_leaf77

You can, but they will rearrange themself such that the composite wavefunction is antisymmetric.

16. Feb 25, 2016

### Happiness

Does it mean forcing the two electrons together will force the wave function of the system to become one that is the same as the one for (d)?

17. Feb 25, 2016

### blue_leaf77

If the degree of freedom is only of the spins, yes it is.

18. Feb 25, 2016

### Happiness

How about I find the composite spin wave function for part (c) by rotating the spin wave function for electron $2$, in the composite spin wave function for part (d), by $\pi$ around the $z$ axis? Then I will get an answer different from the one for part (d). Is there anything wrong?

19. Feb 25, 2016

### Happiness

$\chi=\frac{1}{\sqrt{2}}\alpha_x(1)\beta_x(2)-\frac{1}{\sqrt{2}}\beta_x(1)\alpha_x(2)$

Using $\alpha_x(n)=\frac{1}{\sqrt{2}}\alpha_z(n)+\frac{1}{\sqrt{2}}\beta_z(n)$ and $\beta_x(n)=\frac{1}{\sqrt{2}}\alpha_z(n)-\frac{1}{\sqrt{2}}\beta_z(n)$ and simplfying, I get

$\chi=-\frac{1}{\sqrt{2}}\alpha_z(1)\beta_z(2)+\frac{1}{\sqrt{2}}\beta_z(1)\alpha_z(2)$

which is the same as the one for part (b) but after interchanging electrons $1$ and $2$. Is this expected?

20. Feb 25, 2016

### Happiness

How could we prepare a system with the wave function

$\chi=\frac{1}{\sqrt{2}}\alpha_z(1)\beta_z(2)+\frac{1}{\sqrt{2}}\beta_z(1)\alpha_z(2)$?

21. Feb 25, 2016

### blue_leaf77

You are allowed to do that if the quantum states were just regarded as a vector, i.e. as a mathematical object. But physically, a symmetric state for identical fermions cannot occur (at least until there is an experiment to disprove the symmetrization postulate), therefore your effort is physically meaningless.
I have got the impression that to obtain the last equation, you simply substitute the equations in the second line into the equation in the first line, which implies that you must have arithmetically multiplied the states. That's not allowed for the states are not numbers, they are vector in the relevant vector space. The correct way to express $\chi$ in terms of $\alpha_z\beta_z$ is by regarding them as a composite state, instead of a arithmetic product. I think the easiest way to do this is by working in terms of matrix, e.g. first find the 4x4 matrix for $S_x = S_{1x}+S_{2x}$ in the basis of $\alpha_z\beta_z$ and the find its eigenvectors.
I believe there is no way.

22. Feb 25, 2016

### Happiness

For (c), must it be the case that when being forced together, one electron remains in the "spin up" state in the $x$ direction while the other electron is flipped to the "spin down" state in the $x$ direction? Or could it happen that one electron becomes the "spin up" state while the other electron becomes the "spin down" state, both in the $y$ direction (or in some other direction)?