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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.

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

(2) If not, does this imply there exists a limit on the expected distance between two fermions?

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 $$ N(a) = \Bigg( 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy^1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dx_2a^2+ \mathcal{O}(a^3) \Bigg)^{-1/2} $$$$ N(a) = \Bigg( 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy_1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dx_2a^2+ \mathcal{O}(a^3) \Bigg)^{-1/2} $$ and obtain $$ \Psi(x_1,x_2) = \dfrac{ (\psi(x_1)\psi'(x_2) - \psi(x_2)\psi'(x_1))a + \mathcal{O}(a^2) }{ \sqrt{ 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy^1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dy_2 a^2+ \mathcal{O}(a^3) } } $$$$ \Psi(x_1,x_2) = \dfrac{ (\psi(x_1)\psi'(x_2) - \psi(x_2)\psi'(x_1))a + \mathcal{O}(a^2) }{ \sqrt{ 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy_1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dy_2 a^2+ \mathcal{O}(a^3) } } $$ We evaluate $$ \lim_{a \rightarrow 0} \Psi(x_1,x_2) = \dfrac{ \psi(x_1)\psi'(x_2) - \psi(x_2)\psi'(x_1) }{ \sqrt{ 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy^1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dy_2 } } $$$$ \lim_{a \rightarrow 0} \Psi(x_1,x_2) = \dfrac{ \psi(x_1)\psi'(x_2) - \psi(x_2)\psi'(x_1) }{ \sqrt{ 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy_1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dy_2 } } $$ 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.

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.

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

(2) If not, does this imply there exists a limit on the expected distance between two fermions?

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.

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.

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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.

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

(2) If not, does this imply there exists a limit on the expected distance between two fermions?

General Case

Given a total wavefunction of two fermions with identical spin $$ \Psi(x_1,x_2) = N(a)\Bigg( \psi(x_1)\psi(x_2+a) - \psi(x_2)\psi(x_1+a) \Bigg) $$

one can evaluate its limit towards overlap as

$$ \lim_{a \rightarrow 0} \Psi(x_1,x_2) = \dfrac{ \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_m C_n \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} e^{-\frac{a^2}{2}} a^i \Bigg( \phi_m(x_1)\phi_{i+n}(x_2) - \phi_m(x_2)\phi_{i+n}(x_1) \Bigg) }{\sqrt{2} \sqrt{ 1-e^{-\frac{a^2}{2}}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_n C_{n+i} \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} \left(1-\dfrac{n}{i+1}a+\mathcal{O}(a^2)\right)a^i } } $$ where $\phi_n(x)$ are the eigenfunctions of the quantum harmonic oscillator with unspecified non-vanishing variance and $$ \psi(x) = \sum_{i=0}^{\infty} C_i \phi_i(x) $$we expand with$$ \psi(x_2+a) = \psi(x_2) + \psi'(x_2)a+\mathcal{O}(a^2) $$ $$ \sum_{i=0}^{\infty} (C_i)^2 = 1 $$

Special Case #1

Ifto write $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) \neq 0 $$$$ \Psi(x_1,x_2) = N(a)\Bigg( \psi(x_1) \left(\psi(x_2) + \psi'(x_2)a+\mathcal{O}(a^2)\right) - \psi(x_2) \left(\psi(x_1) + \psi'(x_1)a+\mathcal{O}(a^2)\right) \Bigg) $$ then $\Psi(x_1,x_2)$ is zero in the limit of overlap.

Special Case #2

IfWe reduce $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) = 0 $$$$ \Psi(x_1,x_2) = N(a)\Bigg( \left(\psi(x_1) \psi'(x_2)-\psi(x_2) \psi'(x_1)\right) a + \mathcal{O}(a^2) \Bigg) $$ andWe integrate the square as $$ \dfrac{1}{2} \sum_{n=0}^{\infty} C_n \left( C_n + C_{n+1} \sqrt{(1+n)n} - C_{n+2} \sqrt{(n+2)(n+1)} \right) \neq 0 $$$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}|\Psi(x_1,x_2)|^2dx_1dx_2 \\ = N^2(a) \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Bigg( \left(\psi(x_1) \psi'(x_2)-\psi(x_2) \psi'(x_1)\right) a + \mathcal{O}(a^2) \Bigg)^2 dx_1dx_2 \\ = N^2(a) \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Bigg( \left( (\psi(x_1) \psi'(x_2))^2 + (\psi(x_2) \psi'(x_1))^2 -2 \psi(x_1) \psi'(x_2) \psi(x_2) \psi'(x_1) \right) a^2 + \mathcal{O}(a^3) \Bigg)^2 dx_1dx_2 \\ = N^2(a) \Bigg( 2\int_{-\infty}^{\infty} (\psi(x_1))^2 dx^1 \int_{-\infty}^{\infty}(\psi'(x_2))^2dx_2a^2+ \mathcal{O}(a^3) \Bigg) $$ then $\Psi(x_1,x_2)$ is non-zero and finite in the limit of overlap.

Special Case #3

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) = 0 $$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 $$ \dfrac{1}{2} \sum_{n=0}^{\infty} C_n \left( C_n + C_{n+1} \sqrt{(1+n)n} - C_{n+2} \sqrt{(n+2)(n+1)} \right) = 0 $$$$ \Psi(x_1,x_2) = \dfrac{ (\psi(x_1)\psi'(x_2) - \psi(x_2)\psi'(x_1))a + \mathcal{O}(a^2) }{ \sqrt{ 2\int_{-\infty}^{\infty} (\psi(y_1))^2 dy^1 \int_{-\infty}^{\infty}(\psi'(y_2))^2dy_2 a^2+ \mathcal{O}(a^3) } } $$ then $\Psi(x_1,x_2)$ is divergent inWe 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 ofas two identical fermions approach overlap, they converge to a finite total wavefunction of two centralized orthogonal states.

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. 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.

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

(2) If not, does this imply there exists a limit on the expected distance between two fermions?

General Case

Given a total wavefunction of two fermions with identical spin $$ \Psi(x_1,x_2) = N(a)\Bigg( \psi(x_1)\psi(x_2+a) - \psi(x_2)\psi(x_1+a) \Bigg) $$

one can evaluate its limit towards overlap as

$$ \lim_{a \rightarrow 0} \Psi(x_1,x_2) = \dfrac{ \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_m C_n \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} e^{-\frac{a^2}{2}} a^i \Bigg( \phi_m(x_1)\phi_{i+n}(x_2) - \phi_m(x_2)\phi_{i+n}(x_1) \Bigg) }{\sqrt{2} \sqrt{ 1-e^{-\frac{a^2}{2}}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_n C_{n+i} \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} \left(1-\dfrac{n}{i+1}a+\mathcal{O}(a^2)\right)a^i } } $$ where $\phi_n(x)$ are the eigenfunctions of the quantum harmonic oscillator with unspecified non-vanishing variance and $$ \psi(x) = \sum_{i=0}^{\infty} C_i \phi_i(x) $$ with $$ \sum_{i=0}^{\infty} (C_i)^2 = 1 $$

Special Case #1

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) \neq 0 $$ then $\Psi(x_1,x_2)$ is zero in the limit of overlap.

Special Case #2

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) = 0 $$ and $$ \dfrac{1}{2} \sum_{n=0}^{\infty} C_n \left( C_n + C_{n+1} \sqrt{(1+n)n} - C_{n+2} \sqrt{(n+2)(n+1)} \right) \neq 0 $$ then $\Psi(x_1,x_2)$ is non-zero and finite in the limit of overlap.

Special Case #3

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) = 0 $$ and $$ \dfrac{1}{2} \sum_{n=0}^{\infty} C_n \left( C_n + C_{n+1} \sqrt{(1+n)n} - C_{n+2} \sqrt{(n+2)(n+1)} \right) = 0 $$ then $\Psi(x_1,x_2)$ is divergent in the limit of overlap.

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.

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

(2) If not, does this imply there exists a limit on the expected distance between two fermions?

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.

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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. 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.

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

(2) If not, does this imply there exists a limit on the expected distance between two fermions?

General Case

Given a total wavefunction of two fermions with identical spin $$ \Psi(x_1,x_2) = N(a)\Bigg( \psi(x_1)\psi(x_2+a) - \psi(x_2)\psi(x_1+a) \Bigg) $$

one can evaluate its limit towards overlap as

$$ \lim_{a \rightarrow 0} \Psi(x_1,x_2) = \dfrac{ \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_m C_n \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} e^{-\frac{a^2}{2}} a^i \Bigg( \phi_m(x_1)\phi_{i+n}(x_2) - \phi_m(x_2)\phi_{i+n}(x_1) \Bigg) }{\sqrt{2} \sqrt{ 1-e^{-\frac{a^2}{2}}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_n C_{n+i} \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} \left(1-\dfrac{n}{i+1}a+\mathcal{O}(a^2)\right)a^i } } $$ where $\phi_n(x)$ are the eigenfunctions of the quantum harmonic oscillator with unspecified non-vanishing variance and $$ \psi(x) = \sum_{i=0}^{\infty} C_i \phi_i(x) $$ with $$ \sum_{i=0}^{\infty} (C_i)^2 = 1 $$

Special Case #1

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) \neq 0 $$ then $\Psi(x_1,x_2)$ is zero in the limit of overlap.

Special Case #2

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) = 0 $$ and $$ \dfrac{1}{2} \sum_{n=0}^{\infty} C_n \left( C_n + C_{n+1} \sqrt{(1+n)n} - C_{n+2} \sqrt{(n+2)(n+1)} \right) \neq 0 $$ then $\Psi(x_1,x_2)$ is non-zero and finite in the limit of overlap.

Special Case #3

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) = 0 $$ and $$ \dfrac{1}{2} \sum_{n=0}^{\infty} C_n \left( C_n + C_{n+1} \sqrt{(1+n)n} - C_{n+2} \sqrt{(n+2)(n+1)} \right) = 0 $$ then $\Psi(x_1,x_2)$ is divergent in the limit of overlap.

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. 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.

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

(2) If not, does this imply there exists a limit on the expected distance between two fermions?

General Case

Given a total wavefunction of two fermions with identical spin $$ \Psi(x_1,x_2) = N(a)\Bigg( \psi(x_1)\psi(x_2+a) - \psi(x_2)\psi(x_1+a) \Bigg) $$

one can evaluate its limit towards overlap as

$$ \lim_{a \rightarrow 0} \Psi(x_1,x_2) = \dfrac{ \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_m C_n \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} e^{-\frac{a^2}{2}} a^i \Bigg( \phi_m(x_1)\phi_{i+n}(x_2) - \phi_m(x_2)\phi_{i+n}(x_1) \Bigg) }{\sqrt{2} \sqrt{ 1-e^{-\frac{a^2}{2}}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_n C_{n+i} \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} \left(1-\dfrac{n}{i+1}a+\mathcal{O}(a^2)\right)a^i } } $$ where $\phi_n(x)$ are the eigenfunctions of the quantum harmonic oscillator with unspecified non-vanishing variance and $$ \psi(x) = \sum_{i=0}^{\infty} C_i \phi_i(x) $$ with $$ \sum_{i=0}^{\infty} (C_i)^2 = 1 $$

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) \neq 0 $$ then $\Psi(x_1,x_2)$ is zero in the limit of overlap.

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) = 0 $$ then $\Psi(x_1,x_2)$ is non-zero and finite in the limit of overlap.

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. 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.

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

(2) If not, does this imply there exists a limit on the expected distance between two fermions?

General Case

Given a total wavefunction of two fermions with identical spin $$ \Psi(x_1,x_2) = N(a)\Bigg( \psi(x_1)\psi(x_2+a) - \psi(x_2)\psi(x_1+a) \Bigg) $$

one can evaluate its limit towards overlap as

$$ \lim_{a \rightarrow 0} \Psi(x_1,x_2) = \dfrac{ \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_m C_n \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} e^{-\frac{a^2}{2}} a^i \Bigg( \phi_m(x_1)\phi_{i+n}(x_2) - \phi_m(x_2)\phi_{i+n}(x_1) \Bigg) }{\sqrt{2} \sqrt{ 1-e^{-\frac{a^2}{2}}\sum_{n=0}^{\infty}\sum_{i=0}^{\infty} C_n C_{n+i} \sqrt{\dfrac{n!}{(i+n)!}}\binom {i+n}{n} \left(1-\dfrac{n}{i+1}a+\mathcal{O}(a^2)\right)a^i } } $$ where $\phi_n(x)$ are the eigenfunctions of the quantum harmonic oscillator with unspecified non-vanishing variance and $$ \psi(x) = \sum_{i=0}^{\infty} C_i \phi_i(x) $$ with $$ \sum_{i=0}^{\infty} (C_i)^2 = 1 $$

Special Case #1

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) \neq 0 $$ then $\Psi(x_1,x_2)$ is zero in the limit of overlap.

Special Case #2

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) = 0 $$ and $$ \dfrac{1}{2} \sum_{n=0}^{\infty} C_n \left( C_n + C_{n+1} \sqrt{(1+n)n} - C_{n+2} \sqrt{(n+2)(n+1)} \right) \neq 0 $$ then $\Psi(x_1,x_2)$ is non-zero and finite in the limit of overlap.

Special Case #3

If $$ \sum_{n=0}^{\infty} C_n \left( C_n n - C_{n+1} \sqrt{n+1} \right) = 0 $$ and $$ \dfrac{1}{2} \sum_{n=0}^{\infty} C_n \left( C_n + C_{n+1} \sqrt{(1+n)n} - C_{n+2} \sqrt{(n+2)(n+1)} \right) = 0 $$ then $\Psi(x_1,x_2)$ is divergent in the limit of overlap.

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