Identical particles seem to reduce probability

Question

$\newcommand{\ket}[1]{| #1\rangle}$This question basically has two very related parts. This came up in the context of trying to verify something my professor said a while ago: that if the wave functions for two identical particles are well separated (i.e. if they are very peaked, and the peaks are macroscopically far apart), then you can model them as distinguishable particles. He reasoned that when we swap the wave functions, the term we get from swapping is very small and can be ignored. So, taking the norm of the symmetrized wave function reduces to taking the norm of the non-trivial amplitude in the symmetrization. This will be the amplitude you would get by modeling the particles as distinguishable.

Yet, when I do this procedure, the normalization from the symmetrization procedure screws me up. I am unsure where I am going wrong.

Suppose I have two bosons. I know that one is in the state $\ket{\psi_1}$ and one is in the state $\ket{\psi_2}$.

The symmetrized state then is

$$ \frac{1}{\sqrt{2}} \left(\ket{\psi_1 \psi_2} + \ket{\psi_2\psi_1} \right)$$

Suppose the wave functions for $\ket{\psi_1}$ and $\ket{\psi_2}$ are peaked at separated place, or have non-overlapping support.

Then, if I look at $x_1,x_2$, with $x_1 \in \text{supp}({\psi_1 (\cdot)})$ and $x_2 \in \text{supp}({\psi_2 (\cdot)})$, then won't the probability that I observe a particle near $x_1$ and another near $x_2$ be

$$ \frac{1}{2} |\psi_1(x_1)|^2 |\psi_2(x_2)|^2$$

which is half of what it would be for distinguishable particles? This result feels wrong. I thought maybe something was wrong with the normalization, but it doesn't seem to be that there is.

Answer 1

People often say "quantum indistinguishable particles that are very far-separated behave like distinguishable particles," but this is a bit misleading. It would be more accurate to say "quantum indistinguishable particles that are very far-separated behave like classical indistinguishable particles." The difference is subtle but important.

The problem is in your sentence "if I look at $x_1$, $x_2$, with $x_1 \in \text{supp}(\psi_1(\cdot))$ and $x_2 \in \text{supp}(\psi_2(\cdot))$ ...". There's no possible experiment that says "I've detected particle $x_1$ at location $x$" - only experiments that say "I've detected a particle at location $x$." So the probability amplitude that you measure a particle at point $x$ and another particle at point $y$ is

$$\langle x, y | \frac{1}{\sqrt{2}}(|\psi_1 \psi_2 \rangle + | \psi_2 \psi_1 \rangle) = \frac{1}{\sqrt{2}}(\psi_1(x) \psi_2(y) + \psi_2(x) \psi_1(y))$$

and the actual probability is the norm-squared

$$P(x, y) = \frac{1}{2} \left( |\psi_1(x)|^2 |\psi_2(y)|^2 + |\psi_2(x)|^2 |\psi_1(y)|^2 \right),$$

where the cross-terms are all zero because the states $|\psi_1\rangle$ and $|\psi_2\rangle$ are orthogonal. This is indeed the result for classical indistinguishable particles, and $\int P(x, y)\ dx\, dy = 1$ as it should (unlike your expression).

The only things that differ from the generic quantum case are (a) you can just use the normalization factor of $1/\sqrt{2}$ and (b) when you norm-square the amplitude out to an actual probability density, you can ignore the cross-terms.

If you want to work classically right from the beginning, then you don't symmetrize the ket at all, but work with the ket $|\psi_1 \psi_2 \rangle$ instead. Then your normalization issue doesn't come up.

Answer 2

Instead of writing:

$$ \frac{1}{\sqrt{2}} \left(\ket{\psi_1 \psi_2} + \ket{\psi_2\psi_1} \right)$$

It will be more instructive to see this as

$$ \Psi(x_1, x_2) =\frac{1}{\sqrt{2}} (\psi_1(x_1) \psi_2(x_2) + \psi_1(x_2)\psi_2(x_1)) $$

Thus when you find $|\Psi|^2$, you get three (technically 4) terms:

$$ \frac{1}{{2}} (|\psi_1(x_1) \psi_2(x_2)|^2 + (\text{negligable interaction terms}) + |\psi_1(x_2) \psi_2(x_1)|^2)$$

But for indistinguishable particles, $ \psi_1(x_j) = \psi_1(x_i)$. Although the interaction terms vanish at large distance, this fact holds true regardless of distance between $x_i$ and $x_j$!. Exploiting that fact, and ignoring the interaction terms, we get:

$$ \frac{1}{2}(2 \cdot |\psi_1(x_1) \psi_2(x_2)|^2)$$

And so the result ends up being the same as distinguishable particles, but it just takes a different route to get there.

Edit: Remember that it is $\Psi$ and NOT $\psi$ that must be symmetric/antisymmetric for bosons/fermions respectively under the exchange of particles $x_1$ and $x_2$.

Answer 3

You should use the bosonic symmetrization consistently (with the measured state as well) and say that the probability of detecting one particle in $|\phi_1\rangle$ and one in $|\phi_2\rangle$, where $\langle\phi_1|\phi_2\rangle = 0$, is the magnitude squared of the scalar product between $$|\Psi\rangle = \frac1{\sqrt2} (|\psi_1\rangle|\psi_2\rangle + |\psi_2\rangle|\psi_1\rangle)$$ and $$|\Phi\rangle = \frac1{\sqrt2} (|\phi_1\rangle|\phi_2\rangle + |\phi_2\rangle|\phi_1\rangle).$$ That is, after expansion, the mag-square of $$\langle\Phi|\Psi\rangle = \langle\phi_1|\psi_1\rangle\langle\phi_2|\psi_2\rangle + \langle\phi_2|\psi_1\rangle\langle\phi_1|\psi_2\rangle$$ rather than of $$(\langle\phi_1|\langle\phi_2|)\,|\Psi\rangle = \frac1{\sqrt2}\left(\langle\phi_1|\psi_1\rangle\langle\phi_2|\psi_2\rangle + \langle\phi_2|\psi_1\rangle\langle\phi_1|\psi_2\rangle\right)$$ that you used.

If you bend this formula for use with spatially separated generalized position eigenstates you get the correct result.