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I'm trying to get the wave function of two fermions in a particular case. Writing the Schrödinger equation I can solve it by dividing the wave function into one that depends on the first fermion and one that depends on the second one. $$\Psi(x_1,x_2)=\psi_1(x_1)\psi_2(x_2)$$

If I was trying to solve the case of bosons that would have been great, but because I deal with fermions I somehow need to get that $\Psi(x_1,x_2)=-\Psi(x_2,x_1)$ (and the implied $\Psi(x_1,x_1)=0$) which I don't have.

Any way I tried to antisymmetrize the wave function (mainly using a Slater determinant in different approaches) either gave me 0 or a result that didn't solve the Schrödinger equation.

Is there any other way that can help me antisymmetrize the wave function? If not what does it mean - that this problem has no wave function?

EDIT: Some more data that actually might be important. I'm working in 1+1 dimensions, and in fact the exact solution depends on which particle is to the right, so it's $$ \Psi(x_1,x_2)=\begin{cases} \psi_1(x_1)\psi_2(x_2) & x_2>x_1\\ \psi_1(x_2)\psi_2(x_1) & x_2<x_1 \end{cases} $$ Also I'm only looking for the ground state, so both $\psi_1$ and $\psi_2$ are the ground state solution of each fermion, but I actually have a whole set of solutions if that might help.

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    $\begingroup$ It seems to me that if you need to know which particle is to the right, then the particles can not be identical, and you don't need to antisymmetrize. Perhaps you need to describe the problem in more detail. Or perhaps I'm missing the point. $\endgroup$ – garyp Nov 21 '16 at 17:58
  • $\begingroup$ Either $Ψ(x_1,x_2)=ψ_1(x_1)ψ_2(x_2)−ψ_1(x_2)ψ_2(x_1)$ does the trick or the two fermions are not identical and you don't need to antisymmetrize. $\endgroup$ – Tony May 19 '17 at 14:41
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I suppose that you are not getting the answer because you don't take into account the spin term into the wave function. Look, if you have,

$\psi=\hat{A}\psi={\frac{1}{\sqrt {2}}}\begin{vmatrix} \phi _{1}\alpha & \phi _{1}\beta \\ \phi _{2}\alpha & \phi _{2}\beta \end{vmatrix}={\frac{1}{\sqrt{2}}}[\phi _{1}\phi _{2}(\alpha\beta)-\phi _{1}\phi _{2}(\beta\alpha)]={\frac{1}{\sqrt {2}}}[\phi _{1}\phi _{2}(\alpha\beta-\beta\alpha)](1)$

Where $\phi_{1}$ & $\phi_{2}$ represent respectively, electron 1 & electron 2. $\alpha$ & $\beta$ are the spin orientation, nothing else.

Now, knowing that the fermions (electrons) obey the following equality,

$\psi(x_{1},x_{2})=-\psi(x_{2},x_{1})$

And taking into account Eq. (1)

$\psi(x_{1},x_{2})=-\psi(x_{2},x_{1})=\frac{1}{\sqrt{2}}[\phi _{1}\phi _{2}(\alpha\beta-\beta\alpha)]=-\frac{1}{\sqrt{2}}[\phi _{1}\phi _{2}(\beta\alpha-\alpha\beta)]$

As you may see, there is no 0 as result. Finally if you would want to make the proof about taking the square-integer on $\psi$ you could observe that,

$\psi^{2}=\phi_{1}^{2}\phi_{2}^{2}$

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    $\begingroup$ Please type the text as text and formulas as $\LaTeX$ (MathJax) formulas, not as images. Images are unreadable in many cases, and just don't integrate in the text. For MathJax see tutorial. $\endgroup$ – Ruslan May 19 '17 at 14:56
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This may not be the answer you're looking for, given that you already know about Slater determinants.

For the case of two indistinguishable fermions, the antisymmetrization property (Pauli principle) implies that the total wavefunction must be antisymmetric under particle exchange - i.e. exchange of all respective parameters. Assuming the only internal degree of freedom is spin, there are two cases: (1) spin-state symmetric, position-representation antisymmetric, and (2) spin-state antisymmetric, position-representation symmetric. However if you ignore spin (i.e. assume the spin-state is symmetric and don't consider it any further) then one could write the position-representation of the state as

$$\Psi(x_1,x_2)=\psi_1(x_1)\psi_2(x_2)-\psi_1(x_2)\psi_2(x_1)$$

Notice that if you swap all the $x_1$'s and $x_2$'s you get minus the original wavefunction. This is the Slater determinant with $n=2$ (modulo normalization and up to a phase).

Of course, if you do know that one in particular is on the right, then as @garyp indicated they are no longer indistinguishable and all this goes out the window.

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