Does the many body fermion spatial wavefunction go to zero when two wavefunctions approach each other?

Question

This question assumes the fermions have the same spin eigenstate.

I have been told if somehow one could take the limit as two identical fermion states approach to the same state the total wavefunction goes to zero.

$$ \Psi(1,2) = \big( |1\rangle_1 |2\rangle_2 - |2\rangle_1 |1\rangle_2 \big) $$

However I don't think this is true for continuous spatial wavefunctions.

$$ \Psi(x_1,x_2) = N \left( e^{-\dfrac{|x_1+a|^2}{4 \sigma^2}} e^{-\dfrac{|x_2-a|^2}{4 \sigma^2}} - e^{-\dfrac{|x_2+a|^2}{4 \sigma^2}}e^{-\dfrac{|x_1-a|^2}{4 \sigma^2}} \right) $$

In this example I have two gaussian fermions with equal variance, but there centers are displaced by $2a$. $N$ is the normalization factor which depends on the seperation of the gaussians. I have to solve for $N$ to normalize the total wavefunction to unity.

$$ \Psi(x_1,x_2) = e^{-\dfrac{|x_1|^2+|x_2|^2}{4 \sigma^2}} \dfrac{\sinh\left(\dfrac{a\cdot (x_2-x_1)}{2\sigma^2} \right)\sqrt{2}}{(2\sigma^2\pi)^{\tfrac{n}{2}}\sqrt{\left(e^{\dfrac{a^2}{\sigma^2}}-1\right)}} $$

Above is my total wavefunction when I solve for $N$ and mathematically reduce it.

$$ \lim_{a\rightarrow 0} \Psi(x_1,x_2) = e^{-\dfrac{|x_1|^2+|x_2|^2}{4 \sigma^2}} \dfrac{\hat{a}\cdot (x_2-x_1)}{\sigma(2\sigma^2\pi)^{\tfrac{n}{2}}\sqrt{2}} $$

When I take the limit as the two wavefunctions overlap my total wavefunction converges to a nonzero object.

Does the many body fermion spatial wavefunction go to zero when two wavefunctions approach each other?

General Case

Given a total wavefunction of two fermions with identical spin $$ \Psi(x_1,x_2) = N(a)\Bigg( \psi(x_1)\psi(x_2+a) - \psi(x_2)\psi(x_1+a) \Bigg) $$ we expand $$ \psi(x_2+a) = \psi(x_2) + \psi'(x_2)a+\mathcal{O}(a^2) $$ to write $$ \Psi(x_1,x_2) = N(a)\Bigg( \psi(x_1) \left(\psi(x_2) + \psi'(x_2)a+\mathcal{O}(a^2)\right) - \psi(x_2) \left(\psi(x_1) + \psi'(x_1)a+\mathcal{O}(a^2)\right) \Bigg) $$ We reduce $$ \Psi(x_1,x_2) = N(a)\Bigg( \left(\psi(x_1) \psi'(x_2)-\psi(x_2) \psi'(x_1)\right) a + \mathcal{O}(a^2) \Bigg) $$ We integrate the square as $$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|\Psi(x_1,x_2)|^2dx_1dx_2 \\ = N^2(a) \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Bigg( \left(\psi(x_1) \psi'(x_2)-\psi(x_2) \psi'(x_1)\right) a + \mathcal{O}(a^2) \Bigg)^2 dx_1dx_2 \\ = N^2(a) \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Bigg( \left( (\psi(x_1) \psi'(x_2))^2 + (\psi(x_2) \psi'(x_1))^2 -2 \psi(x_1) \psi'(x_2) \psi(x_2) \psi'(x_1) \right) a^2 + \mathcal{O}(a^3) \Bigg)^2 dx_1dx_2 \\ = N^2(a) \Bigg( 2\int_{-\infty}^{\infty} (\psi(x_1))^2 dx^1 \int_{-\infty}^{\infty}(\psi'(x_2))^2dx_2a^2+ \mathcal{O}(a^3) \Bigg) $$ . We solve for $$ N(a) = \Bigg( 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy_1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dx_2a^2+ \mathcal{O}(a^3) \Bigg)^{-1/2} $$ and obtain $$ \Psi(x_1,x_2) = \dfrac{ (\psi(x_1)\psi'(x_2) - \psi(x_2)\psi'(x_1))a + \mathcal{O}(a^2) }{ \sqrt{ 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy_1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dy_2 a^2+ \mathcal{O}(a^3) } } $$ We evaluate $$ \lim_{a \rightarrow 0} \Psi(x_1,x_2) = \dfrac{ \psi(x_1)\psi'(x_2) - \psi(x_2)\psi'(x_1) }{ \sqrt{ 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy_1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dy_2 } } $$ This proves that if you take the continuum limit as two identical fermions approach overlap, they converge to a finite total wavefunction of two centralized orthogonal states.

Answer 1

Pretty much by construction your limit is non-zero: you normalize it all the way as $a\to0$. If you simply consider the norm of your wavefunction as a function of $a$, you'll see that it's $1$ for all $a>0.$ Then the limit is trivially also $1$, which is non-zero and proves your point.

I have been told if somehow one could take the limit as two identical fermion states approach to the same state the total wavefunction goes to zero.

This of course only works if you don't try to re-normalize it under the limit to prevent from going to zero.

Answer 2

I think that the error you are making is that the limit of $a->0$ may not exist in the sense you think it does. What you are trying to do is to normalize a wave function that has zero norm.

In all cases where $a\neq 0$ what you are doing is mathematically sound, and yields an unambiguous result. I'm quite sure though that the point $a=0$ may not have a waeel defined limit as you might get different result depending on with direction of $a$ you are approaching from. The fact that you get a $\hat a\cdot(x_1-x_2)$ in you final result is an indication of that.

Physically what goes wrong is simply that before you normalize $\psi_{a=0}(x_1,x_2)=0$ identically.

Answer 3

You are misunderstanding the Pauli Exclusion Principle (PEP). It states that two Fermions cannot occupy the same state. That means they cannot be represented by the exact same wave function. A wave function for a Fermion can be separable into different parts and the PEP does not exclude some of those parts being identical. For example, the wave function for Fermions consists of a spatially dependent part and a spin dependent part. Hence the spatially dependent wave functions can be identical and yet the PEP is not violated as long as the spin wave function components are not identical.

Update: Now that some errors and misunderstandings have been cleared away (see the comments below), the issue of what relationship this finding has to the PEP can be addressed. The two-body wave function being investigated is in the form of a Slater determinant. The Slater determinant was introduced originally in the context of approximate solutions to the Many-Body Schrodinger equation. Slater noted that the simple product of single particle wave functions could violate the PEP under certain conditions and, furthermore, did not reflect the indistinguishability of identical quantum particles. He introduced the determinantal form (this is how he referred to it in his lectures) to overcome these deficiencies. This form insures that any choice of single particle wave functions that violates the PEP will automatically yield a value of 0 for the Many-Body wave function.

One aspect of Many-Body wave functions of determinantal form is that they do not require normalization as long as the single-particle wave functions themselves are orthonormal. This is not the case for the SP wave functions chosen here, but as pointed out by @MichaelFremling, the second "renormalization" employed in this analysis to overcome the fact that the two-body wave function vanishes as a approaches 0 (just as Slater intended) probably does not yield a well defined limit for the two-body wave function as a goes to 0. In effect the analysis yields a finite nonzero result by multiplying 0 by infinity, but this limit could be manipulated to yield any result whatsoever. For this reason, this analysis has no relationship to the PEP. I hesitate to say that it has no relationship to quantum mechanics, because the determinantal form does reflect an entanglement between the single-particle wave functions. The method employed here of studying the effect of separation of otherwise identical wave functions on a Many-Body system may have some implications for the study of entanglement. I'll leave it for others to explore this possibility.

One final point is that the determinantal form is not an exact treatment of a Many-Body wave function. It was introduced as an improved approximation in a Many-Body solution that replaced the real two-body interactions by a mean field potential. The corrections to this approximation are known as correlation interactions. This form does include the exchange correlation (what differentiates Hartree-Fock from Hartree approximation), however, the PEP is itself a form of correlation that is probably not fully represented by the determinantal form.

Despite my negative conclusions of what this means for the PEP, I can state that (for me) this has been a very thought provoking question and I encourage further research along these lines.

Answer 4

You wrote the two-particle state as

$$| \Psi \rangle = \big( |1\rangle_1 |2\rangle_2 - |2\rangle_1 |1\rangle_2 \big).$$

Assuming the single-particle states $| 1 \rangle$ and $| 2 \rangle$ are normalized, then this two-particle wavefunction is only correctly normalized if they are orthogonal. In general we have $$\langle \Psi | \Psi \rangle = \big( \langle 1 |_1 \langle 2 |_2 - \langle 2 |_1 \langle 1 |_2 \big) \big( |1\rangle_1 |2\rangle_2 - |2\rangle_1 |1\rangle_2 \big) \\ = \langle 1 | 1 \rangle \langle 2 | 2 \rangle - \langle 1 | 2 \rangle \langle 2 | 1 \rangle - \langle 2 | 1 \rangle \langle 1 | 2 \rangle + \langle 2 | 2 \rangle \langle 1 | 1 \rangle \\ = 2 \left( 1 - |\langle 1 | 2 \rangle|^2 \right)$$

So the correctly normalized two-particle state is actually

$$| \Psi \rangle = \frac{1}{\sqrt{2 \left( 1 - |\langle 1 | 2 \rangle|^2 \right)}} \big( |1\rangle_1 |2\rangle_2 - |2\rangle_1 |1\rangle_2 \big).$$

(This is just the more general version of the calculation that you performed.) If you were to take the two-particle wavefunction that you originally defined - one which is antisymmetrized but not normalized - then it would indeed go to zero in the limit where "the two wavefunctions approach each other," i.e. $\langle 1 | 2 \rangle \to 1$. But this fact is not particularly interesting or important, because a physically useful wavefunction should be normalized. The properly normalized wavefunction $| \Psi \rangle$ does not go to zero because the normalization constant diverges to infinity at just the right rate to keep the wavefunction normalizable.

The Pauli Exclusion Principle is tremendously physically important, but it doesn't refer to the process you are considering of moving the two entire wavefunctions together. Instead, it refers to a situation where the two-body wavefunction is fixed, and says that the probability amplitude of finding the two particles very close together and thus $P(x,x)$ go to zero. Consider the probability of finding the particles and positions $x$ and $y$; this is given by

$$ P(x,y) = \Big| \big( \langle x |_1 \langle y |_2 \big) | \Psi \rangle \Big|^2 = \mathcal{N}^2 \left| \langle x | 1 \rangle \langle y | 2 \rangle - \langle x | 2 \rangle \langle y | 1 \rangle \right|^2.$$

Now the point is that the wavefunction $| \Psi \rangle$, and therefore the normalization constant $\mathcal{N}$, stays fixed as you take the limit $y \to x$, and so the probability amplitude vanishes. However, in this case you're not actually changing the state $| \Psi \rangle$ or the two-particle wavefunction itself.