I want to calculate the probability to find two bosons at the same place.

Let the bosonic wave function be $$\Psi_B(x_1,x_2)=\frac{1}{\sqrt{2}}\left(\Psi_1(x_1)\Psi_2(x_2)+\Psi_2(x_1)\Psi_1(x_2)\right).$$

Then I have to calculate the expectation value of $\delta(x_1-x_2):$ $$\begin{align}\langle\delta(x_1-x_2)\rangle_{\Psi_B}&=\int_{\mathbb{R}^3}\int_{\mathbb{R}^3}\overline{\Psi_B(x_1,x_2)}\delta(x_1-x_2)\Psi_B(x_1,x_2)d^3x_2d^3x_1\\ &=\frac{1}{2}\int_{\mathbb{R}^6}\overline{2\Psi_1(x_1)\Psi_2(x)}\cdot2\Psi_1(x_1)\Psi_2(x)d^6x\\ &=2||\Psi_1\Psi_2||^2=2>1.\end{align}$$

This result is obviously wrong. Why is that, though?

  • $\begingroup$ First of all, when integrating over the delta function, one of the intregrals should vanish. Secondly, in the last line you should still have an integral left, since you will be integrating $|\psi_1(x_1)|^2|\psi_2(x_2)|^2$ this is not nececarilly one. $\endgroup$ – Mikael Fremling Jul 26 '17 at 13:05
  • $\begingroup$ So let's say I first integrate over $d^3x_2$. Then I am left with an integral over $d^3x_1$ and the same integrand, except with $x_2=x_1$, i.e. $\int_{\mathbb{R}^3}\overline{\Psi_B(x_1,x_1)}\Psi_B(x_1,x_1)d^3x_1$? $\endgroup$ – Thomas Wening Jul 26 '17 at 13:24
  • $\begingroup$ Exacly. And this will not be $=1$ in general. $\endgroup$ – Mikael Fremling Jul 26 '17 at 13:37
  • $\begingroup$ Ah, now I understand! But is there a way to show that this is generally less than 1? $\endgroup$ – Thomas Wening Jul 26 '17 at 13:39

This probability is actually $0$. This is a 2d variant of the question: "What is the probability of find the system described by $\psi(x)$ at position $x_0$?". In both cases the answer is $0$. In 1d this is because the probability is the area under a single point of height $\vert \psi(x_0)\vert^2$, and this point has no width so the area is $0$. In the 2d case of this question, the answer if $0$ because this probability is a line of finite length but no width, so the area under this line is $0$.

This does not mean that finding the expectation value of $\delta(x_1-x_2)$ cannot be done. It's just that $\langle\delta(x_1-x_2)\rangle$ does not give the probability you are looking for.

The same situation occurs in 1d. $\langle \delta (x)\rangle$ simply returns $\vert\psi(0)\vert^2$, which incidentally can be $>1$: there is nothing to prevent $\psi(x)$ or $\vert\psi(x)\vert^2$ to be $>1$ since the probability is the probability density $\vert\psi(x)\vert^2$ integrated over an interval. For small intervals $dx$ we have $\vert\psi(x)\vert^2dx<1$. An easy example of this is the normalized wavefunction $\psi(x)=\sqrt{2}\frac{e^{-2x^2}}{\pi^{1/4}}$. Then $$ \langle \delta(0)\rangle = \int dx\, \frac{2e^{-4x^2}}{\sqrt{\pi}}\delta(0)= \frac{2}{\sqrt{\pi}}\approx 1.13. $$

Now, in the 2d case, to find $\langle\delta(x_1-x_2)\rangle$, you obtained quite correctly the expectation value of $\delta(x_1-x_2)$ as \begin{align} \langle\delta(x_1-x_2)\rangle&= \int dx_1 \int dx_2 \delta(x_1-x_2) \textstyle\frac{1}{2} \vert \Psi_1(x_1)\Psi_2(x_2)+\Psi_2(x_1)\Psi_1(x_2)\vert^2\, ,\\ &=\int dx_1 \textstyle\frac{1}{2}\vert \Psi_1(x_1)\Psi_2(x_1)+\Psi_2(x_1)\Psi_1(x_1)\vert^2\, ,\\ \end{align} At this point, the computation continues as

\begin{align} &=\int dx_1 \textstyle\frac{1}{2}\vert \Psi_1(x_1)\Psi_2(x_1)+\Psi_2(x_1)\Psi_1(x_1)\vert^2\, ,\\ &=\int dx_1 \textstyle\frac{1}{2} \vert2 \Psi_1(x_1)\Psi_2(x_1)\vert^2\, ,\\ &=2\int dx_1 \vert \Psi_1(x_1)\Psi_2(x_1)\vert^2\, ,\\ &=2\int dx_1 \vert \Psi_1(x_1)\vert^2\vert \Psi_2(x_1)\vert^2 \, , \tag{1}\\ &\ne 2 \int dx_1 \vert \Psi_1(x_1)\vert^2\int dx_1 \vert \Psi_2(x_1)\vert^2\, . \end{align}

Of course the factor of 2 in Eq.(1) is expected since bosons are more likely to bunch together than independent particles, in other words the probability amplitude $\Psi(x,x)$ for two bosons at the same place is enhanced by $\sqrt{2}$ over the probability amplitude for distinguishable particles.

As a calculational example consider $\psi_0(x)$ and $\psi_1(x)$ to be harmonic oscillator wavefunctions: $$ \psi_0(x)=\frac{e^{-\frac{x^2}{2}}}{\sqrt[4]{\pi }}\, ,\qquad \psi_1(x)=\frac{\sqrt{2} e^{-\frac{x^2}{2}} x}{\sqrt[4]{\pi }} $$ The probability density is then $$ \vert\Psi_B(x,y)\vert^2=\frac{e^{-x^2-y^2} \left(-2 x^2 y-x \left(2 y^2-1\right)+y\right)^2}{2 \pi } $$ Setting $x=y$ we obtain $$ \vert\Psi_B(x,x)\vert^2=\frac{2 e^{-2 x^2} x^2 \left(1-2 x^2\right)^2}{\pi } $$ and \begin{align} \int dx \vert\Psi_B(x,x)\vert^2&=\frac{7}{8\sqrt{2\pi}}\approx 0.35\, ,\\ &=2\int dx \vert\psi_0(x)\vert^2 \vert\psi_1(x)\vert^2 \end{align}


To calulate the the probability to find two bosons at the same place to need to evaluate

$$\begin{align}\langle\delta(x_1-x_2)\rangle_{\Psi_B}&=\int_{\mathbb{R}^3}|{\Psi_B(x,x)}|^2d^3x\\ &=\int_{\mathbb{R}^3}|\Psi_1(x)|^2|\Psi_2(x)|^2d^3x.\end{align}$$

This is clearly smaller than 1 since you are multiplying together two normalized probability distributes and integrating over them.

  • 1
    $\begingroup$ The problem is that if you set $x_1=x_2=x$ in your $\Psi_B(x,x)$ you get $\Psi_B(x,x)=\sqrt{2}\Psi_1(x)\Psi_2(x)$, not $\Psi_1(x)\Psi_2(x)$. $\endgroup$ – ZeroTheHero Jul 26 '17 at 20:12

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