4
$\begingroup$

In quantum mechanics, particles are described as wave-like. This means, for example, that an electron or photon does not have a well-defined position before one measures it and causes the wavefunction to collapse. Instead, the particle has a probability distribution of where it is likely to be found when measuring.

Now I wondered if the spin of a particle behaves similarly. For example, could an electron have a spin of $\frac 1 2$ and "at the same time", a spin of $-\frac 1 2$? (By "at the same time", I mean that one cannot say for sure what its spin is - similar to position where one maybe could say that the particle either has no position or is everywhere at the same time).

If yes, how could this be in agreement with the Pauli exclusion principle? (If two electrons have undefined spin, how does for example an atomic orbital "know" that it is filled and no third electron could enter?)

$\endgroup$
4
  • 1
    $\begingroup$ I have edited out the word 'rate' from the title, as it opens a completely different discussion. Quantum particles don't have spin "rates" -- spin is a form of angular momentum, but it doesn't imply anything is "spinning". For more details se e.g. this question, or have a deeper search for previous discussions on this site. $\endgroup$ Dec 18 '20 at 16:58
  • 1
    $\begingroup$ Superpositions also work with spins. An electron can be in the $\left( \left| +\frac{1}{2} \right\rangle + \left| -\frac{1}{2} \right\rangle \right) / \sqrt{2}$ state. If you add another electron in the same place, it needs to be of the opposite spin: $\left( \left| +\frac{1}{2} \right\rangle_1 \left| -\frac{1}{2} \right\rangle_2 - \left| -\frac{1}{2} \right\rangle_1 \left| +\frac{1}{2} \right\rangle_2 \right) / \sqrt{2}$. $\endgroup$
    – A. P.
    Dec 18 '20 at 17:00
  • $\begingroup$ From the title I had thought it was the entanglement of let say a spin 1 and a spin 1/2 particle, were it a fermion or a boson, as it has to be considered as a whole (unseparability) ? $\endgroup$
    – Cretin2
    Dec 18 '20 at 17:10
  • $\begingroup$ @JulienPitteloud I meant the electron being in spin superposition as described by A. P. $\endgroup$
    – Jonas
    Dec 18 '20 at 17:14
6
$\begingroup$

Let's take this in steps:

  • The phrase "at the same time" is shorthand for quantum superposition.
  • "Having spin 1/2" means two different things: having total spin number 1/2 (which means that the total spin angular momentum has a well-defined value of $|\mathbf S|^2 = \hbar^2 s(s+1)$ for $s=1/2$) and having a well-defined spin projection $S_z=\frac12 \hbar$ (i.e., the $z$ component of the spin angular momentum vector $\mathbf S$). As a general rule, the total spin quantum number $s$ of a given particle is fixed (and can never be negative), whereas the spin projection $m_s$ can change depending on the situation, from $-s$ to $s$ in integer steps. For $s=1/2$, $m_s$ can be either $1/2$ or $-1/2$.
  • It is perfectly possible for the spin of a quantum particle to be in a superposition state of different spin projections, such as the $m_s=1/2$-superposed-with-$m_s=-1/2$ configuration that you mentioned.
  • In a quantum superposition, which we generally denote in a form like $$a\left|\tfrac12\right> + e^{i\phi}b\left|-\tfrac12\right>,$$ we care about the relative amplitudes $a$ and $b$ of the two states (whose squares give the relative probabilities for the particle to be found in those states) as well as their relative phase $e^{i\phi}$, a complex number (often just equal to $+1$ or $-1$) that encodes the quantum nature of the combination.
  • This quantum superposition state can be encoded as a complex-valued vector $(a,e^{i\phi}b)\in\mathbb C^2$. Two states encoded by $(a_1,e^{i\phi_1}b_1)$ and $(a_2,e^{i\phi_2}b_2)$ can be considered to be completely distinguishable if their inner product is zero, i.e., if $$\left<(a_1,e^{i\phi_1}b_1),(a_1,e^{i\phi_1}b_1)\right>= a_1a_2+e^{i(\phi_1-\phi_2)}b_1b_2 = 0$$ (assuming real $a_i$ and $b_i$).
  • In these terms, the Pauli exclusion principle is compatible with electrons being in superposition states, but the states of two electrons need to be completely distinguishable. As an example, an electron in the superposition state $$\left|\tfrac12\right> + \left|-\tfrac12\right>$$ can be in the same place (i.e. same spatial distribution) as another electron in the 'opposite' superposition state $$\left|\tfrac12\right> - \left|-\tfrac12\right>,$$ but no other electrons will fit into that spatial state.

More generally, it is important to emphasize that a superposition state is not, in any way, an "undefined spin". It is a very clearly defined state, which just happens not to have a well-defined projection along the chosen $z$ axis.

If you want to dig deeper, the "grown-up" version of the Pauli exclusion principle is the formalism for indistinguishable quantum particles, which requires (for fermions) that the wavefunction change sign if two fermions are exchanged. This has wide-ranging consequences and is explained in detail in intermediate QM textbooks .One particular consequence is the "smaller" version of the Pauli exclusion principle, that electrons cannot be in the same state $-$ as then the global wavefunction would be symmetric instead of antisymmetric.

$\endgroup$

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.