Let consider a fermionic $2$-particles system (eg: $\text{H}_2$ or Helium atom). It's total antisymmetrical (!) wave function $\psi(\vec{r_1}, s_1,\vec{r_2}, s_2)$ can be expressed as the product of the spatial and spin wave function:

$$\psi\left(\vec{r}_1, s_1,\vec{r}_2, s_2\right)= \psi_{\text{orb}}\left(\vec{r}_1,\vec{r}_2\right) \times \psi_{\text{spin}}(s_1,s_2)$$

where $\psi_{\text{orb}}$ is the spatial WF and $\psi_{\text{spin}}$ the spin WF. That is in order that the total WF is antisym, then we have two cases: $\psi_{\text{orb}}$ is symmetric (s) and $\psi_{\text{spin}}$ antisymmetric (a) or vice versa.

I have a question on the nomeclature for the spin WF:

It is known that the two-particle spin WF is represented in the basis that consists of the eigenvectors of the operators $\hat{S}^2$ and $\hat{S}_z$ where $\hat{S}=s_1 +s_2$. These are:

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with antisymm singlet ($S=0$) and symm triplet ($S=1$)

My Question is: Is it true that by convention it is synonymous to say

$\psi_{\text{spin}}$ is in state $S=1$ and that the both spins are "parallel" to each other and

resp. that $\psi_{\text{spin}}$ is in state $S=0$ and that the both spins are "anti parallel" to each other?

Background: The characterisations that classify anti- and ferromagnetism using the exchange integral

$$J_{ex}= E_s - E_a= 2\langle \,\psi _{A}({\vec r}_{1})\psi _{B}({\vec r}_{2})\,|V|\,\psi _{A}({\vec r}_{2})\psi _{B}({\vec r}_{1}),\rangle$$

($E_s$ is the eigenenergy of symmetric spatial WF and $E_a$ " anti symmetrical " )

use the terminology that spins are parallel and antiparallel to each other in dependence if $J >0$ or $J <0$, but geometrically I not understand it.

What I see is that since $J >0$ or $J <0$ allows to conclude if spinal WF is in singlet or triplet state, but the usage of "parallel" and "antiparallel" iritates me at this point, I came to a guess that "parallel" is by convention set to be synonymous to triplet state and antiparallel to singlet.

Is my guess correct or do I miss the issue?


No. Parallel corresponds to $$S_{z1}S_{z2}|\psi\rangle=\left(\frac{\hbar}{2}\right)^2|\psi\rangle$$ And anti-parallel corresponds to $$S_{z1}S_{z2}|\psi\rangle=0$$ Thus $|\uparrow\uparrow\rangle$ and $|\downarrow\downarrow\rangle$ are parallel states and the other two are antiparallel.

  • $\begingroup$ This is exactly the point that confuses me so much. Geometrically, your explanation make sense and coinsides exactly with my intuition before I read this: en.wikipedia.org/wiki/Exchange_interaction#Effects_of_exchange There the exchnage factor $J_{ab}= \frac{1}{2}(E_+-E_-)$ where $E_+$ is the eigenvalue of symmetric spatial wave function $\psi_{orb}^s$ and $E_-$ is the eigenvalue of antisymmetric spatial wave function $\psi_{orb}^a$. Now the cruical point: according to my knowledge the system tends to be in a state of lowest energy. That gives me two cases: $\endgroup$ – KarlPeter Jan 22 '20 at 11:50
  • $\begingroup$ 1. $J_{ab} >0$: this implies $E_+ > E_-$. Therefore the system tends to have a spatial WF corresponding to $E_-$: so the spatial WF is antisymmetric. As electrons are fermions and therefore it's total WF is antisymm, the spin WF $\psi_{spin}$ is symmetrical, thus a triplet state $S=1$. That's fine. But this gives no informations on $m_S$. And thus I don't see why in case $J_{ab} >0$ the spins are parallel as claimed in linked wiki exerpt. $\endgroup$ – KarlPeter Jan 22 '20 at 11:50
  • $\begingroup$ 2. Analogously, in case for $J_{ab} <0$ going similar argumentation using same arguments I would be able to conclude that the spin WF has to be singulet: $S=0$. But again, wki says that $J_{ab} <0$ imply that the spins are antiparallel. Why? I don't understand exactly these two points. What is my thinking error? $\endgroup$ – KarlPeter Jan 22 '20 at 11:50
  • $\begingroup$ The Hamiltonian for a (ferro)magnetic system is generally given by spin-spin interactions between nearest neighbours of a lattice structure. So it’s of the form $J_{ij}S_i\cdot S_j$ And this is why the parallel and anti parallel come into the picture. $\endgroup$ – Superfast Jellyfish Jan 22 '20 at 13:01
  • $\begingroup$ I see that due to the shape of the Heisenberg Hamiltonian ${\displaystyle {\mathcal {H}}_{Heis}=-\sum _{i,j}J\langle {\vec {S}}_{i}\cdot {\vec {S}}_{j}\rangle }$ the parallel and antiparall spins are 'somehow' related to $J_{ab}$. But I still not see a clear logical argument why e.g. $J_{ab}>0$ imply that spins are parallel (resp. $J_{ab}<0$ that spins $\endgroup$ – KarlPeter Jan 22 '20 at 13:35

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