The left-right symmetry and forbidden triplet spin state

[hokhani]

TL;DR

The symmetry and two-spin state

Suppose that we have a system including two single (and identical) orbitals as shown in the attached image. Then, consider the states that we have two different spins on the two orbitals. It seems that because of the left right symmetry of the system the special part of the wave function can not be asymmetric and so we can not have triplet (symmetric) spin wave function and only the singlet state is permitted. I would like to know if this reasoning is correct. Any help is appreciated.

 

[PeterDonis]

@hokhani, please note that putting equations and text in images is not permitted. Please use the PF LaTeX feature to post equations directly. There is a "LaTeX Guide" link at the bottom left of the post window.

 

[PeterDonis]

consider the states that we have two different spins on the two orbitals

If you stipulate that the two electron spins must have opposite values (i.e., total spin zero), that already rules out two of the three triplet states. But on what grounds are you making this stipulation?

It seems that because of the left right symmetry of the system the special part of the wave function can not be asymmetric

I assume you mean "spatial" (not "special"). It is not correct that left right symmetry of the system (meaning its Hamiltonian) requires that the spatial wave function must be symmetric. You can have spatially asymmetric wave functions in this case; they will just occur in pairs, with each one of a pair being the mirror image of the other.

 

[hokhani]

If you stipulate that the two electron spins must have opposite values (i.e., total spin zero), that already rules out two of the three triplet states. But on what grounds are you making this stipulation?

The opposite spins does't necessarily mean that the total spin is zero. We can also have the state $$\frac{1}{\sqrt(2)}(\uparrow \downarrow+\downarrow \uparrow)$$.

I assume you mean "spatial" (not "special"). It is not correct that left right symmetry of the system (meaning its Hamiltonian) requires that the spatial wave function must be symmetric. You can have spatially asymmetric wave functions in this case; they will just occur in pairs, with each one of a pair being the mirror image of the other.

Right, sorry for typo. Could you please write the asymmetric spatial wavefunction in this case?

 

[PeroK]

The opposite spins does't necessarily mean that the total spin is zero. We can also have the state $$\frac{1}{\sqrt(2)}(\uparrow \downarrow+\downarrow \uparrow)$$.

Right, sorry for typo. Could you please write the asymmetric spatial wavefunction in this case?

The overall wavefunction for a system of two identical fermions must be antisymmetric. If the spatial wavefunction is symmetric, then the spin "component" of the wavefunction must be antisymmetric. That implies the singlet state. This applies in the particular case when the fermions share a spatial wavefunction, such as the same orbital in an atom - in which case the spin state must be the singlet.

 

[PeterDonis]

Could you please write the asymmetric spatial wavefunction in this case?

You wrote one possibility already (although it has an index error in it, which I'll correct), the "minus" part of your spatial $\psi$ wave function:

$$
\psi_\text{spatial} = \psi_\alpha (x_1) \psi_\beta (x_2) - \psi_\alpha(x_2) \psi_\beta (x_1)
$$

This is antisymmetric, so it's one wave function of a pair that are mirror images of each other. The other is what you get if you swap $x_1$ and $x_2$.

 

[vanhees71]

The state
$$\frac{1}{\sqrt{2}} (|1/2,-1/2\rangle+|-1/2,1/2\rangle )$$
belongs to total spin $S=1$. It's the state with $\Sigma=0$. The other two basis vectors with $\Sigma=\pm 1$ are $|1/2,1/2 \rangle$ and $|-\1/2,-1/2 \rangle$, respectively.

The state with $S=\Sigma=0$ is
$$\frac{1}{\sqrt{2}} (|1/2,-1/2\rangle-|-1/2,1/2\rangle ).$$
If you have the spin part in the triplet (singlet) state this state is symmetric (anti-symmetric) under exchange, the spatial part must be anti-symmetric (symmetric).

 

[hokhani]

You wrote one possibility already (although it has an index error in it, which I'll correct), the "minus" part of your spatial $\psi$ wave function:

$$
\psi_\text{spatial} = \psi_\alpha (x_1) \psi_\beta (x_2) - \psi_\alpha(x_2) \psi_\beta (x_1)
$$

This is antisymmetric, so it's one wave function of a pair that are mirror images of each other. The other is what you get if you swap $x_1$ and $x_2$.

Alright, this wave function is zero because of symmetry and so we only have the symmetric spatial wave function (with singlet spin wave).

 

[PeterDonis]

this wave function is zero because of symmetry

Why? As I've already said, left-right symmetry does not mean the spatial wave function has to be symmetric. It just means that antisymmetric wave functions have to occur in pairs, with each one the left-right mirror image of the other. There is nothing physically impossible about the antisymmetric spatial wave functions; they just imply that if the electrons are in that spatial state, their spin wave function must be symmetric.

I suspect that what you are actually trying to find is the ground state, i.e., the state of lowest energy, for this system. But if so, you should state that explicitly, because finding the ground state is a much more restrictive problem than just finding states that can have nonzero probability in general.

 

[vanhees71]

Take helium as an example. There you have "para helium" and "ortho helium". One are those in the spin singulet and triplet states of its two electrons. The ground state is a para helium state, because you get a much lower energy for the spatially symmetric wave functions.

https://en.wikipedia.org/wiki/Helium_atom

 

[hokhani]

Why? As I've already said, left-right symmetry does not mean the spatial wave function has to be symmetric. It just means that antisymmetric wave functions have to occur in pairs, with each one the left-right mirror image of the other. There is nothing physically impossible about the antisymmetric spatial wave functions; they just imply that if the electrons are in that spatial state, their spin wave function must be symmetric.

I suspect that what you are actually trying to find is the ground state, i.e., the state of lowest energy, for this system. But if so, you should state that explicitly, because finding the ground state is a much more restrictive problem than just finding states that can have nonzero probability in general.

I pointed that we are talking about the specific case which two electrons with opposite spins are on two orbitals.
In this case, because of symmetry we have $x_1=-x_2$ and $\alpha=\beta$ and we can infer that the spatial part is zero.

 

[PeterDonis]

I pointed that we are talking about the specific case which two electrons with opposite spins are on two orbitals.

Yes. So what?

In this case, because of symmetry we have $x_1=-x_2$

No, we don't. The fact that the electrons are in the left and right orbitals does not mean their $x$ coordinates must satisfy this relation. The orbitals are not points.

and $\alpha=\beta$

This makes no sense. $\alpha$ and $\beta$ are labeling the two electrons. They can't both be the same electron.

 

[vanhees71]

What we are discussing is best described by a tensor product of the form $\text{spatial part} \otimes \text{spin part}$, which can be done in non-relativistic approximation, where the spin commutes with position and momentum operators. For a system of two spin-1/2 particles as discussed here, there are two possibilities:
$$\Psi(\vec{x}_1,\sigma_1;\vec{x}_2 \sigma_2)=\psi_{\text{even}}(\vec{x}_1,\vec{x}_2) \otimes \chi^{(S=0)}(\sigma_1,\sigma_2)$$
or
$$\Psi(\vec{x}_1,\sigma_2;x_2 \sigma_2) = \psi_{\text{odd}}(\vec{x}_1,\vec{x}_2) \otimes \chi^{(S=1)}(\sigma_1,\sigma_2).$$
In the first case there's only one spin state,
$$\chi^{(S=0)}(\sigma_1,\sigma_2)=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle - |-1/2,1/2 \rangle),$$
while in the 2nd case
$$\chi^{(S=1})(\sigma_1,\sigma_2) = \lambda_1 |1/2,1/2 \rangle + \lambda_0 \frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle + |-1/2,1/2 \rangle_ + \lambda_{-1} |-1/2,-1/2 \rangle.$$
Since we must have
$$\Psi(\vec{x}_1,\sigma_1;\vec{x}_2,\sigma_2)=-\Psi(\vec{x}_2,\sigma_2;\vec{x}_1,\sigma_1),$$
obviously we must have
$$\psi_{\text{even}}(\vec{x}_1,\vec{x}_2)=\psi_{\text{even}}(\vec{x}_2,\vec{x}_1), \quad \psi_{\text{odd}}(\vec{x}_1,\vec{x}_2)=-\psi_{\text{odd}}(\vec{x}_2,\vec{x}_1),$$
because
$$\chi^{(S=0)}(\sigma_1,\sigma_2)=-\chi^{(S=0)}(\sigma_2,\sigma_1), \quad \chi^{(S=1)}(\sigma_1,\sigma_2) = \chi^{(S=1)}(\sigma_2,\sigma_1).$$

 

[hokhani]

Yes. So what?No, we don't. The fact that the electrons are in the left and right orbitals does not mean their $x$ coordinates must satisfy this relation. The orbitals are not points.

I attached a non-symmetric configuration of electrons in the orbital in the mirror-symmetric system. Do you mean that we can have such an electronic position?

This makes no sense. $\alpha$ and $\beta$ are labeling the two electrons. They can't both be the same electron.

 

[DrClaude]

$$
\psi_\text{spatial} = \psi_\alpha (x_1) \psi_\beta (x_2) - \psi_\alpha(x_2) \psi_\beta (x_1)
$$

Alright, this wave function is zero because of symmetry and so we only have the symmetric spatial wave function (with singlet spin wave).

$\psi_\text{spatial}$ is only zero when $x_1=x_2$. The only thing you can say is that the probability of finding both electrons at the same place is zero.

 

[PeterDonis]

I attached a non-symmetric configuration of electrons in the orbital in the mirror-symmetric system. Do you mean that we can have such an electronic position?

You are misunderstanding my point. Even in the "symmetric" configuration of orbitals you originally drew, the fact that electron 1 is in the left orbital while electron 2 is in the right orbital does not mean that $x_1 = - x_2$. The orbitals are not points, they are regions of space, so they don't correspond to just one value of $x$.

 

[hokhani]

You are misunderstanding my point. Even in the "symmetric" configuration of orbitals you originally drew, the fact that electron 1 is in the left orbital while electron 2 is in the right orbital does not mean that $x_1 = - x_2$. The orbitals are not points, they are regions of space, so they don't correspond to just one value of $x$.

Orbitals are nothing except for wave functions which are functions of $x$. Therefore, I think my reasoning runs again in this regard.

 

[PeterDonis]

Orbitals are nothing except for wave functions which are functions of $x$.

Sure. But they're not delta functions; they aren't zero except for a single value of $x$.

Therefore, I think my reasoning runs again in this regard.

You have given no argument whatever to support your reasoning that, if electron 1 is in the left orbital and electron 2 is in the right orbital, we must have $x_1 = - x_2$. And that claim is obviously false if either of the orbitals contains more than one point, i.e., is nonzero for more than one value of $x$. As of course both are.

 

[hokhani]

Sure. But they're not delta functions; they aren't zero except for a single value of $x$.You have given no argument whatever to support your reasoning that, if electron 1 is in the left orbital and electron 2 is in the right orbital, we must have $x_1 = - x_2$. And that claim is obviously false if either of the orbitals contains more than one point, i.e., is nonzero for more than one value of $x$. As of course both are.

My reasoning is based on the mirror symmetry of the system. In the figure, I sent in my previous post, could you please explain the possibility of having instantly the two electrons in the non symmetric configuration while the system is symmetric?

 

[PeterDonis]

My reasoning is based on the mirror symmetry of the system.

No, it isn't. It's based on this invalid claim of yours:

I pointed that we are talking about the specific case which two electrons with opposite spins are on two orbitals.
In this case, because of symmetry we have $x_1=-x_2$ and $\alpha=\beta$ and we can infer that the spatial part is zero.

The claim that "because of symmetry we have $x_1 = - x_2$" is false. I have already explained why.

The claim that because of symmetry $\alpha = \beta$ is also false, as I have already pointed out; but one false claim is already enough to invalidate your argument, so we don't need another.

 

[PeterDonis]

In the figure, I sent in my previous post, could you please explain the possibility of having instantly the two electrons in the non symmetric configuration while the system is symmetric?

The fact that "the system is symmetric" does not mean the exact electron positions have to be symmetric all the time. "The system is symmetric" just means that the Hamiltonian of the system is invariant under parity inversion about the "mirror" surface shown in your drawings.

In fact, saying that one electron is in the left orbital and one electron is in the right orbital does not imply a definite position (value of $x$) for either electron at all. The orbitals, as I have already pointed out, are not delta functions. So the whole way you are writing down the spatial wave functions is not really correct to begin with.

 

[hokhani]

The fact that "the system is symmetric" does not mean the exact electron positions have to be symmetric all the time. "The system is symmetric" just means that the Hamiltonian of the system is invariant under parity inversion about the "mirror" surface shown in your drawings.

In fact, saying that one electron is in the left orbital and one electron is in the right orbital does not imply a definite position (value of $x$) for either electron at all. The orbitals, as I have already pointed out, are not delta functions. So the whole way you are writing down the spatial wave functions is not really correct to begin with.

Thank you very much for pointing the parity inversion concept which seems that resolved my confusion about the symmetric system. As I underestood by your comment, the mirror symmetry means that $H(x_1,x_2)=H(-x_1,-x_2)$ which doesnt demand the condition $x_1=-x_2$.