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I am trying to formulate the wave function that describes two entangled electons having the same position but opposite spin. According to the Pauli exclusion principle this should be possible. And when their spin would be equal, the wave function would cancel out and become 0.

I tried: $\psi({r}_1,r_2,s_1,s_2)=\psi_a (r_1, s_1) \psi_b (r_2, s_2) - \psi_b (r_1, s_1) \psi_a (r_2, s_2)$.

With $r_1=r_2=: r$ and $s_2=s_1 =: s$ (equal spin states) this becomes: $\psi({r}_1,r_2,s_1,s_2)=\psi_a (r, s) \psi_b (r, s) - \psi_b (r, s) \psi_a (r, s)=0$.

With $r_1=r_2=: r$ and $s_2=-s_1 =: -s$ (opposite spin states) this becomes: $\psi({r}_1,r_2,s_1,s_2)=\psi_a (r, s) \psi_b (r, -s) - \psi_b (r, s) \psi_a (r, -s)$.

But via the spin property: $\psi (r, -s)= -\psi (r, s)$ I get $0$ again:
$\psi({r}_1,r_2,s_1,s_2)= -\psi_a (r, s) \psi_b (r, s) + \psi_b (r, s) \psi_a (r, s) = 0$.

What is/are the mistake(s)?

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What is the "spin property" you mention? –  Robin Ekman Jul 9 at 1:54
    
The "spin property" as I understand it, is that when $\psi$ is the wave function of an electron with some position, energy, spin direction etc. then the wave function of the same electron with opposite spin direction must be multiplied with -1. –  Gerard Jul 9 at 15:11
    
Where did you read this? It is not true. The state representing a spin in the opposite direction is orthogonal to the original state. A multiplication by $-1$ represents the same state. –  Robin Ekman Jul 9 at 15:14

2 Answers 2

up vote 1 down vote accepted

$\newcommand{\spinup}{\vert{\uparrow\rangle}} \newcommand{\spindown}{\vert{\downarrow\rangle}}$

The "spin property" you mention is false. Since spin up and spin down by definition have different eigenvalues of the spin operator, they must be orthogonal states. Therefore for a general state, there is no relation between $\psi(r,s)$ and $\psi(r,-s)$ other than normalization. In classical physics, you have of course that the component along $s$ is minus the component along $-s$, but in quantum mechanics $\psi(r,s)$ is not the component of the spin vector along $s$, but the probability amplitude to detect the particle in a state with spin $s$. Other than that the particle has some spin (the total probability is $1$) there can be no general relation between the amplitudes for $s$ and $-s$.

The simplest way to understand states with spin is, I think, that for a single particle with spin, $\psi(x)$ is vector-valued with one component for each spin state. More concretely for a spin-$\frac 1 2$ particle, $$\psi(x) = \alpha(x)\spinup + \beta(x)\spindown$$ with $$\int |\alpha(x)|^2 + |\beta(x)|^2 \, dx = 1$$ for normalization. This is because the probability to be found in an interval $dx$ with spin up is $|\alpha(x)|^2 \, dx$, and in the same interval with spin down, $|\beta(x)|^2 \, dx$, so the integrand is the probability to be found in the interval $dx$, regardless of spin.

Consider the states $$\psi_a(x)= \alpha(x) \spinup \qquad \psi_b(x)=\alpha(x)\spindown $$ which have opposite spin, but the same distribution in space: both particles are found in an interval $dx$ with probability $|\alpha(x)|^2 \, dx$. The antisymmetric combination of these states is \begin{align} \Psi(x,y) & = \alpha(x)\alpha(y)\spinup\spindown -\alpha(y)\alpha(x)\spindown\spinup \\ & = \alpha(x)\alpha(y) \big( \spinup\spindown - \spindown\spinup \big). \end{align} The reason that this does not vanish identically is that $$\spinup\spindown - \spindown\spinup \neq 0.$$ So it is possible to have a state symmetric in space if the spins are opposite. This is your second calculation with the spin part of the wavefunction written out.

If instead we take $$\psi_a(x) = \alpha(x)\spinup \qquad \psi_b(x) = \beta(x)\spinup$$ we will find that the antisymmetric combination is $$\Psi(x,y) = (\alpha(x)\beta(y) - \beta(x)\alpha(y))\spinup\spinup.$$ As you can see when $x = y$, the probability to find equal spins is indeed $0$. (Since the calculation with spin down is identical.) This is your first calculation.

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Thank you for taking the time to explain. –  Gerard Jul 9 at 17:03

I think the "spin property" you are quoting is not applicable (no idea where it comes from). If you have one electron with spin projection $+\frac{1}{2}$, then $\psi(r, +\frac{1}{2})$ would be non-zero for some r, but $\psi(r, -\frac{1}{2})$ would be zero everywhere.

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