Spin wave function of a system of 2 electrons

[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?

 

[DrClaude]

(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 $\chi=\frac{1}{\sqrt{2}}\alpha(1)-\frac{1}{\sqrt{2}}\beta(1)$?

Conventionally, the one with a + sign.

(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 $\chi=\frac{1}{\sqrt{2}}\alpha(1)\beta(2)-\frac{1}{\sqrt{2}}\beta(1)\alpha(2)$?

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

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

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

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

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

 

[blue_leaf77]

χ=α(1)β(2)χ=α(1)β(2)\chi=\alpha(1)\beta(2). The others are entangled states.

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?

 

[DrClaude]

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?

Yes, I was negligent.

 

[Happiness]

Yes, I was negligent.

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 $\chi=\frac{1}{\sqrt{2}}\alpha(1)+\frac{1}{\sqrt{2}}\beta(1)$?

 

[Happiness]

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

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?

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

 

[DrClaude]

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

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

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

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.

 

[Happiness]

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

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

 

[blue_leaf77]

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

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 $\chi=\frac{1}{\sqrt{2}}\alpha(1)\beta(2)-\frac{1}{\sqrt{2}}\beta(1)\alpha(2)$.

Since the spin wave function is independent of the azimuthal angle ϕϕ\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?

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)..

 

[Happiness]

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.

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?

 

[blue_leaf77]

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?

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.

 

[Happiness]

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.

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.

 

[blue_leaf77]

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

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

 

[Happiness]

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

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

 

[blue_leaf77]

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

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

 

[Happiness]

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

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)?

 

[blue_leaf77]

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)?

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

 

[Happiness]

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

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?

 

[Happiness]

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.

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

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?

 

[Happiness]

How could we prepare a system with the wave function

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

 

[blue_leaf77]

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?

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.

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

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?

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.

How could we prepare a system with the wave function

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

I believe there is no way.

 

[Happiness]

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

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)?