-1
$\begingroup$

Quoting from Eisberg Resnick Quantum Physics:

If we consider space variables of two electrons (identical particles) to have almost the same values, then their wavefunctions are 'almost' identical if they are in the same quantum state, ie, $\psi_{a}(1)~ \simeq~\psi_{a}(2)$ and $\psi_{b}(1)~\simeq~\psi_{b}(2)$ [the label 1 and 2 denote the spatial co-ordinates of the electron '1' and '2' i.e. ($x_1,y_1,z_1$) and ($x_2,y_2,z_2$), and the labels a and b for the wavefunction denote the three quantum numbers $n,l,m$ of two different quantum states].

In this case, the antisymmetric space eigenfunction describing the system of two electrons is

$$\frac{1}{\sqrt2}[\psi_{a}(1)\psi_{b}(2) - \psi_{a}(2) \psi_{b}(1)]\simeq\frac{1}{\sqrt2}[\psi_{b}(1)\psi_{a}(2) - \psi_{b}(1) \psi_{a}(2)]\simeq 0$$

and the symmetric space eigenfunction is

$$\frac{1}{\sqrt2}[\psi_{a}(1)\psi_{b}(2) + \psi_{a}(2) \psi_{b}(1)]\simeq\frac{1}{\sqrt2}[\psi_{b}(1)\psi_{a}(2) + \psi_{b}(1) \psi_{a}(2)]\simeq\sqrt2\psi_{b}(1) \psi_{a}(2). $$

So, my question is why is it that we observe 2 electrons to have parallel spin(triplet) and antisymmetric space wavefunction (whose probability density is zero) (just satisfying pauli's exclusion principle here), as compared to antiparallel spins (singlet) and symmetric spatial wavefunction? I was reading up on ferromagnetism on this wiki article and also i had asked this question about a month back in chemistry stack exchange here which wasnt really well recieved.

JUST understand this; i completely get the idea of less coulumb repulsion when two electrons are located further apart which favours triplet state. What i dont understand is that we have calculated the probability density of finding the two electrons to be in almost the same spatial state for symmetric spatial wavefunction as $$|\sqrt2\psi_{b}(1) \psi_{a}(2)|^2$$

which is more than that of the antisymmetric spatial wavefunction (whose probability density is nearly zero)........what does this mean? More probability density but not found that way in nature at all? Again, i understand that less coulumb repulsion means lesser energy ...but what about the probability density?


This whole thing can be very confusing; let me try and clear it up a little more. There are two kinds of situation at play here.

Electrons having same spatial co-ordinates (and this state is further subdivided into 2 kinds....namely symmetric and antisymmetric spatial wavefunction)

Electrons having different spatial co-ordinates (again same subdivision...symmetric and antisymmetric spatial wavefunction)

$\endgroup$
1
$\begingroup$

Let $\psi_A({\bf r}_1,{\bf r}_2)$ be the antisymmetric spatial wavefunction. It vanishes at locations where ${\bf r}_1 = {\bf r}_2$. However, it is still normalised overall. Therefore its vanishing along the line ${\bf r}_1 = {\bf r}_2$ is compensated by its being extra-large elsewhere, at places where ${\bf r}_1 \ne {\bf r}_2$. Overall, therefore, the situation of electrons adopting the antisymmetric spatial wavefunction is not an unlikely situation. What your question amounts to is the observation that when the spin state is symmetric (sometimes called 'parallel' spins, though I prefer to avoid that terminology), which implies the spatial state must be $\psi_A$, the electrons move around in a correlated way, such that they tend not to approach one another closely. This is sometimes called a 'Fermi hole' in the joint probability distribution. If we force a pair of electrons to approach one another closely, by introducing hard walls and squashing them in, then the antisymmetric spatial state is still available, along with other states. Owing to Coulomb repulsion it will typically be a state of lower energy than the spatially symmetric alternative composed of the same wavefunctions but with a plus sign. This does not, on its own, dictate any probability of preferring one state or the other. However, since many physical processes such as photon emission and thermal relaxation tend to favour moving towards lower energy states, the one with lower energy is more often occupied, other things being equal. Having said that, it often happens that other contributions to the energy result in the ground state being one in which both electrons have the same spatial wavefunction, which automatically results in a symmetric spatial state and therefore the singlet spin state. This will happen, for example, when the kinetic energy contributes more to the energy overall than does the Coulomb repulsion between the electrons. Atoms in group 2 give examples.

$\endgroup$
1
$\begingroup$

why is it that we observe 2 electrons to have parallel spin(triplet) and antisymmetric space wavefunction (whose probability density is zero) (just satisfying pauli's exclusion principle here), as compared to antiparallel spins (singlet) and symmetric spatial wavefunction?

Actually, both situations exist and are commonly found in nature. For example, a helium atom has two electrons; atoms with their electrons in the former configuration are called "orthohelium" and atoms with their electrons in the latter state are called "parahelium." As you said, orthohelium excited states have lower energy because the electron separation lowers the repulsive Coulomb energy, and is therefore more common, but both states definitely exist.

$\endgroup$
  • $\begingroup$ Yes i know ortho and parahelium; but are those the cases which pertain to the electrons having 'almost' the same spatial co-ordinates? My question was more about that. By the way if you can help me reach another conclusion, ground state of helium is a singlet state as given here and do we call this parahelium? or 1 electron in 1s orbital and the other in 2s orbital is where we have the nomenclature ortho and para for parallel and antiparallel spins respectively? . $\endgroup$ – Prasad Mani Nov 13 '16 at 3:56
  • $\begingroup$ May want to look at this. Either way, i dont think ortho and para helium deal with electrons being in same spatial co-ordinates and then debate over the possibility of it having symmetric or antisymmetric wavefunction $\endgroup$ – Prasad Mani Nov 13 '16 at 3:57
  • $\begingroup$ **debate over the possibility of it having symmetric or antisymmetric SPATIAL wavefunction $\endgroup$ – Prasad Mani Nov 13 '16 at 4:08
0
$\begingroup$

Focusing on same spatial part is a red herring I think. The point is that if the part of non-spin wavefunction is anti-symmetric the spin part will be triplet and vice versa. The system will most likely be in the state with greater probability. Hence the two configurations for helium.

Now coming to your question, if you could confine the system in a small volume (to overcome coulomb repulsion) then the system would be found in the singlet state as it has higher probability. However if you set the system free then the electrons would be pushed apart. You could maybe do an adiabatic calculation to see how probabilities change. Also remember that at the cross over point the transition still has to obey angular momentum conservation so if you think the system goes from singlet to triplet you'd have to account for the angular momentum change in electron system. If the transition point is quite far can a photon be (non-locally) absorbed to bring in the required extra angular momentum?

$\endgroup$
0
$\begingroup$

The two electron wave function must be antisymmetric under particle exchange. Either the spatial part of the wave function is antisymmetric and the spin part is symmetric, or vice versa. Singlet spin functions are antisymmetric so the must combine with a symmetric space function. For triplet states it is the other way around.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.