Confusion In Writing Identical Particle Wavefunctions

[flyusx]

Homework Statement

See below.

Relevant Equations

Wavefunction With Spatial ($\phi$) And Spin ($\chi$) Components: $\psi=\phi\chi$
Bosons: Symmetric Wavefunctions
Fermions: Antisymmetric Wavefunctions

I had read up on identical particles and their associated symmetric/antisymmetric wavefunctions a while back and solved a few problems. It seems like I'm still confused on some fronts. I have picked here two solved problems from Zettili's QM book (Edition 3) that I believe illustrate the part I do not understand.

Problem 9.16(a) – Two identical particles of spin-1/2 are enclosed in a one-dimensional box potential of length $L$ with walls at $x=0$ and $x=L$. Find the energies and of the three lowest states.

This is really a perturbation theory problem in the subsequent parts, but Zettili starts by writing down the system's wavefunction and this is where I got confused. His first sentence is "Since the two particles have the same spin, the spin wave function of the system must be symmetric, so $\chi_{s}$ is any one of the triplet states"...and then he tags along an antisymmetric spatial wavefunction to produce an overall antisymmetric wavefunction. Why did he eliminate the possibility of writing a wavefunction with a symmetric spatial part with the antisymmetric spin singlet state?

When I re-read his chapter on identical particles, I found Problem 8.4 for which I have the same confusion; I've decided to include this problem as well since it may illustrate my thought process a bit better.

Problem 8.4 – Neglecting the spin–orbit interaction and the interaction between the electrons, find the energy levels and the wave functions of the three lowest states for a two-electron atom.

My confusion arises when he finds the wavefunction for the first excited state. Since one electron occupies $n=1$ and the other $n=2$, he writes the solution $$\psi=\frac{1}{\sqrt{2}}\left(\phi_{1,0,0}\left(r_{1}\right)\phi_{2,0,0}\left(r_{2}\right)-\phi_{2,0,0}\left(r_{1}\right)\phi_{1,0,0}\left(r_{2}\right)\right)\chi_{\text{triplet}}$$ which I understand because the spatial part is antisymmetric and the triplet state is symmetric. But I have two concerns here:

1. Why is the solution $$\frac{1}{\sqrt{2}}\left(\phi_{1,0,0}\left(r_{1}\right)\phi_{2,0,0}\left(r_{2}\right)+\phi_{1,0,0}\left(r_{2}\right)\phi_{2,0,0}\left(r_{1}\right)\right)\chi_{\text{singlet}}$$ not present? Permuting $r_{1},r_{2}$ shows that the spatial component is symmetric which combined with the antisymmetric singlet state gives an antisymmetric wavefunction.
2. This is a bit of a sidenote, but my intuition tells me that I should I be able to replace $\phi_{2,0,0}$ with any of the three other $n=2$ states (that all correspond to $l=1$) because their energies are degenerate when ignoring finer interactions.

I suppose my question can be summed up as 'why are not all possible (in my eyes) antisymmetric wavefunctions really possible'? Thanks in advance!

 

[vela]

Problem 9.16(a) – Two identical particles of spin-1/2 are enclosed in a one-dimensional box potential of length $L$ with walls at $x=0$ and $x=L$. Find the energies and of the three lowest states.

This is really a perturbation theory problem in the subsequent parts, but Zettili starts by writing down the system's wavefunction and this is where I got confused. His first sentence is "Since the two particles have the same spin, the spin wave function of the system must be symmetric, so $\chi_{s}$ is any one of the triplet states"...and then he tags along an antisymmetric spatial wavefunction to produce an overall antisymmetric wavefunction. Why did he eliminate the possibility of writing a wavefunction with a symmetric spatial part with the antisymmetric spin singlet state?

I don't get it either. I think he's wrong, but the same text is in the second edition. I would think that an obvious error like that would have been caught and corrected already, so perhaps we're both missing something here.

My confusion arises when he finds the wavefunction for the first excited state. Since one electron occupies $n=1$ and the other $n=2$, he writes the solution $$\psi=\frac{1}{\sqrt{2}}\left(\phi_{1,0,0}\left(r_{1}\right)\phi_{2,0,0}\left(r_{2}\right)-\phi_{2,0,0}\left(r_{1}\right)\phi_{1,0,0}\left(r_{2}\right)\right)\chi_{\text{triplet}}$$ which I understand because the spatial part is antisymmetric and the triplet state is symmetric. But I have two concerns here:

1. Why is the solution $$\frac{1}{\sqrt{2}}\left(\phi_{1,0,0}\left(r_{1}\right)\phi_{2,0,0}\left(r_{2}\right)+\phi_{1,0,0}\left(r_{2}\right)\phi_{2,0,0}\left(r_{1}\right)\right)\chi_{\text{singlet}}$$ not present? Permuting $r_{1},r_{2}$ shows that the spatial component is symmetric which combined with the antisymmetric singlet state gives an antisymmetric wavefunction.
2. This is a bit of a sidenote, but my intuition tells me that I should I be able to replace $\phi_{2,0,0}$ with any of the three other $n=2$ states (that all correspond to $l=1$) because their energies are degenerate when ignoring finer interactions.

I had the same thoughts reading through his solution. It seems he should have mentioned the points you brought up to avoid confusion on the student's part.

 

[flyusx]

I don't get it either. I think he's wrong, but the same text is in the second edition. I would think that an obvious error like that would have been caught and corrected already, so perhaps we're both missing something here.


I had the same thoughts reading through his solution. It seems he should have mentioned the points you brought up to avoid confusion on the student's part.

Thanks for your reply. I had wrote down the additional solutions when I originally solved it, but I thought I may have been wrong after being confused in the 9th chapter. It's nice to get some additional confirmation.

 

[PeroK]

His first sentence is "Since the two particles have the same spin, the spin wave function of the system must be symmetric, so $\chi_{s}$ is any one of the triplet states"...and then he tags along an antisymmetric spatial wavefunction to produce an overall antisymmetric wavefunction. Why did he eliminate the possibility of writing a wavefunction with a symmetric spatial part with the antisymmetric spin singlet state?

This I really don't get. The reason that two electrons may occupy the ground state of an atom is precsiely because they may be in the (antisymmetric) singlet spin state. It's not that different from two particles in a box, as far as I can see.

 

[kuruman]

This I really don't get. The reason that two electrons may occupy the ground state of an atom is precsiely because they may be in the (antisymmetric) singlet spin state. It's not that different from two particles in a box, as far as I can see.

This has me puzzled as well. The only thing I can think of is that, in the problem description, by "same spin" the author means "both spin up" which can only be if they are in the triplet state, i.e. the spin-symmetric wavefunction is assumed to be the case a priori. It's a strange way of formulating the problem.

 

[vela]

This has me puzzled as well. The only thing I can think of is that, in the problem description, by "same spin" the author means "both spin up" which can only be if they are in the triplet state, i.e. the spin-symmetric wavefunction is assumed to be the case a priori. It's a strange way of formulating the problem.

The same thought occurred to me; however, the solution explicitly lists all three of the triplet states and claims there's a threefold degeneracy of the ground state. It's clearly admitting the possibility of the two electrons having opposite spin.

 

[kuruman]

The same thought occurred to me; however, the solution explicitly lists all three of the triplet states and claims there's a threefold degeneracy of the ground state. It's clearly admitting the possibility of the two electrons having opposite spin.

I thought that the threefold degeneracy is in the three $|m_s\rangle$ states of the triplet in the absence of the spin-orbit interaction.

 

[vela]

I thought that the threefold degeneracy is in the three $|m_s\rangle$ states of the triplet in the absence of the spin-orbit interaction.

It is. My point was that if same spin meant both spin-up or both spin-down, it would correspond to only two of the triplet states and the solution wouldn't have admitted all three states and a threefold degeneracy.

 

[kuruman]

After thinking about this some more, I think we need more information from the OP. I believe we all agree that the three lowest energy levels are as shown schematically on the right. The OP wonders why the answer includes the excited triplet but does not include the excited singlet.

To @flyusx: What does Zettili say the second excited state is instead of the singlet?

This hyperphysics link shows the helium energy levels and can be used for a reality check.

 

[flyusx]

View attachment 364223After thinking about this some more, I think we need more information from the OP. I believe we all agree that the three lowest energy levels are as shown schematically on the right. The OP wonders why the answer includes the excited triplet but does not include the excited singlet.

To @flyusx: What does Zettili say the second excited state is instead of the singlet?

This hyperphysics link shows the helium energy levels and can be used for a reality check.

Hey, sorry for the late reply! Zettili says the second excited state is $\phi_{2,0,0}\left(r_{1}\right)\phi_{2,0,0}\left(r_{2}\right)\chi_{\text{singlet}}$.

If it's of any help, he says the ground state energy is $2E_{1}=-27.2Z^{2}$ eV, the first excited energy is $E_{1}+E_{2}=-17Z^{2}$ eV and the second excited is $2E_{2}=-6.8Z^{2}$ eV. I suppose his problem means the three lowest distinct energies?

Then he says 'these reuslts are obviously not expected to be accurate because, by neglecting the Coulomb interaction between the electrons, we have made a grossly inaccurate approximation'. The theoretical ground state energy from this method is -108.8 eV while the experimental value is -78.975 eV.

 

[kuruman]

I suppose his problem means the three lowest distinct energies?

Yes.

The theoretical ground state energy from this method is -108.8 eV while the experimental value is -78.975 eV.

In the ground state the two electrons have the same spatial distribution which kinda means that they are on top of one another. If you include the electron-electron repulsion term, you expect the atom to be less tightly bound, i.e. increased ground state energy.

 

[PeroK]

If I were studying this, I would file this away. Even if there is something we're missing it's not necessarily critical. That's a risk. The bigger risk is that you waste previous time.

Make a mental note and move on. That's my advice.

 

[flyusx]

If I were studying this, I would file this away. Even if there is something we're missing it's not necessarily critical. That's a risk. The bigger risk is that you waste previous time.

Make a mental note and move on. That's my advice.

I agree. I wrote a note in the book margins about including the singlet state a while back and have reached time-independent perturbation theory now. I just wanted to double check. Thanks!