3
$\begingroup$

I have two electrons which are in a (symmetric) triplet state where one of them has spin pointing up ($m_s=1/2$) and the other one has spin pointing down ($m_s=-1/2$), so we have

$|1,0\rangle =\;(\uparrow\downarrow + \downarrow\uparrow)/\sqrt2$

According to Pauli's exclusion principle, the two electrons must not have the same set of quantum numbers. However, as the two spin directions differ, is it possible that the corresponding spatial wave functions for the electrons are both in ground state ($n=1,l=0,m_l=0$)? Or is it a contradiction to the fact that it is a triplet state?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

The spin wave function is symmetric with respect to the exchange of particles. Therefore the spacial wave function has to be antisymmetric. I.e. at least one of the quantum numbers has to be different.

The wave function may look as if the electrons have opposite spin, but actually the spins are the same if measured at an axis 90° from z.

EDIT:

The spin eigenvectors of different axes are not independent from each other. $$ \begin{align} \left|\strut\uparrow\right\rangle &= \frac{1}{\sqrt 2}\left(\left|\strut x_+\right\rangle + \left|\strut x_-\right\rangle \right)\\ \left|\strut\downarrow\right\rangle &= \frac{1}{\sqrt 2}\left(\left|\strut x_+\right\rangle - \left|\strut x_-\right\rangle \right) \end{align} $$ Substitute that into your definition of $\left|\strut\psi\right\rangle$ and you will get $$ \begin{align} \left|\strut\psi\right\rangle &= \frac{1}{2\sqrt 2}\left[\left(\left|\strut x_+\right\rangle + \left|\strut x_-\right\rangle \right)\otimes\left(\left|\strut x_+\right\rangle - \left|\strut x_-\right\rangle \right) + \left(\left|\strut x_+\right\rangle - \left|\strut x_-\right\rangle \right)\otimes\left(\left|\strut x_+\right\rangle + \left|\strut x_-\right\rangle \right)\right]\\ &=\frac{1}{2\sqrt 2}\left[% \left|\strut x_+\right\rangle\left|\strut x_+\right\rangle% - \left|\strut x_+\right\rangle\left|\strut x_-\right\rangle% + \left|\strut x_-\right\rangle\left|\strut x_+\right\rangle% - \left|\strut x_-\right\rangle\left|\strut x_-\right\rangle% + \left|\strut x_+\right\rangle\left|\strut x_+\right\rangle% + \left|\strut x_+\right\rangle\left|\strut x_-\right\rangle% - \left|\strut x_-\right\rangle\left|\strut x_+\right\rangle% - \left|\strut x_-\right\rangle\left|\strut x_-\right\rangle% \right]\\ &=\frac{1}{2\sqrt 2}\left[% \left|\strut x_+\right\rangle\left|\strut x_+\right\rangle% - \left|\strut x_-\right\rangle\left|\strut x_-\right\rangle% + \left|\strut x_+\right\rangle\left|\strut x_+\right\rangle% - \left|\strut x_-\right\rangle\left|\strut x_-\right\rangle% \right]\\ &=\frac{1}{\sqrt 2}\left[% \left|\strut x_+\right\rangle\left|\strut x_+\right\rangle% - \left|\strut x_-\right\rangle\left|\strut x_-\right\rangle% \right]\\ \end{align} $$

To check for the symmetry you don't need this calculation. It is sufficient to check that $$ \frac{1}{\sqrt2}\left( \left|\strut\uparrow\right\rangle\left|\strut\downarrow\right\rangle + \left|\strut\downarrow\right\rangle\left|\strut\uparrow\right\rangle \right) = \frac{1}{\sqrt2}\left( \left|\strut\downarrow\right\rangle\left|\strut\uparrow\right\rangle + \left|\strut\uparrow\right\rangle\left|\strut\downarrow\right\rangle \right) $$

The Pauli principle says that wave functions have to be negated when swapping any two Fermions.

Improve this answer
$\endgroup$
4
  • $\begingroup$ How do you see they have the same spin in a certain direction? More precisely which direction in the x-y plane are you referring to? I find it hard to belive because the reduced density matrix of each electron is maximally mixed and I would think that that's a basis independent statement. $\endgroup$
    – Borun Chowdhury
    Commented Jan 7, 2017 at 8:40
  • $\begingroup$ OK, so if the quantum numbers of the spatial wave function would be identical, there would be no way of getting this function asymmetric. To the question where I can see the direction: well, I can see that one has spin-up (so the z-component is pointing upwards) and one has spin-down (z is pointing downwards) $\endgroup$
    – user141316
    Commented Jan 7, 2017 at 12:37
  • $\begingroup$ Thanks for the calculation, in my book it didn't go into much detail about this. So if we have an electron with a set of quantum numbers ($n=1,l=0, m_l=0, s=1/2, m_s=1/2$), we cannot say if another electron with the config $n=1,l=0, m_l=0, s=1/2, m_s=-1/2$ exists, unless we know that the spin is asymmetric, right? $\endgroup$
    – user141316
    Commented Jan 7, 2017 at 12:50
  • $\begingroup$ Often people just describe a multi particle state as if it were composed of distinct individual particles. If they say one electron has spin up and the other has spin down ($\left|\strut\uparrow\downarrow\right\rangle$) what they are actually talking about is $S_-\left|\strut\uparrow\downarrow\right\rangle$ which is $\frac{1}{\sqrt2}\left(\left|\strut\uparrow\downarrow\right\rangle-\left|\strut\downarrow\uparrow\right\rangle\right)$ $\endgroup$
    – Leonard Michlmayr
    Commented Jan 7, 2017 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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