For some reason, there have always been some aspects of Quantum Mechanics which I’ve found it hard to get to grips with, not for the usual reasons ( weirdness, macroscopic superposition &c), but simply because some of the basic mathematical consequences of the framework are neither obvious nor typically explored well in the literature. Most of these puzzles first hit me when dealing with identical particles, and also, separately, with the framework of field theory. So this note is my attempt to get to grips with things in a way I can refer back to the next time I get puzzled.
The first thing which puzzled me related to boson states. We are told that if we have a hamiltonian \(H1\) and a state space \(V1\) which represent a single particle, that the tensor product \(V1 \otimes V1 \otimes \ldots \)… represents \(n\) non-interacting distinguishable particles, and that the subspace of this space where the states are symmetrized represents the corresponding valid states for non-interacting bosons. So for example, in a two-state system, two bosons occupy a three dimensional space, and two of those three dimensions correspond to having the two particles in the same state. Thus, when doing statistical mechanics, the likelihood of finding the particles in the same state is greater than it would be if the particles were distinguishable.
So far so good. But now we are told something different: if we have \(n\) bosons in the same state and one boson in a different state, and we evolve the system in some way, the likelihood of the other boson joining the first \(n\) is enhanced by a factor of \(n+1\). So in our two state system with two bosons, we should be twice as likely to see both bosons evolve into the same state as would be the case if the particles were non-identical. And with a million bosons, we’d be incrementing a probability by a factor of a million.
This caused me great confusion. What was there to stop the probability of the appropriate transition of distinguishable particles being greater than one in a million, thus leading to a nonsensical probability greater than one.? After all, I though, it’s easy to construct a unitary matrix whose transition probabilities are as close as we like to unity. Well, so it is, but although the unitary matrix connecting the single particle states can be anything unitary we like, the matrix connecting the multi-particle states is a tensor product of unitary matrices, and that, it turns out, makes all the difference. In fact, the distinguishable particle probability for \(n\) particles is always safely below \(1/n\) because of the consequence of multiplying \(n\) factors from a unitary matrix together to get the required amplitude.
Let’s actually do a calculation. Start off by assuming that we have two states of interest and \(n\) particles. \(n-1\) of them are currently in state \(|1>\) , and the other in state \(|2>\) .
Now lets assume we have a unitary matrix \(U\) representing some kind of event, perhaps a complex interaction, perhaps just the passage of time. Then the transition probability we want is \( |(a1^n a2)|^2 \) where the first factor represents the amplitude \( a1 = <1|U|1> \) for the particles already in state 1 to remain there, and the second factor \( a2 = <1|U|2>\) the amplitude for the other particle to join them.
Now, the only thing we know about \(a1\) and \(a2\) is that \(|a1|^2 + |a2|^2 <= 1\) since certainly \(U\) is unitary. The inequality is because there might be other states into which a transition can occur, as well as the two of interest.
For notational convenience, define \( p = |a1|^2 \) so that \( |a2|^2 < 1-p \) and our probability \( P < p^{n} (1-p) \)
So we are now in the realms of a simple mathematical question: if
$$ f(p) = p^n * (1-p) $$
then does \(p< 1\) imply \(f(p) < \frac 1 n \)
Now \( f(0) = f(1) = 0 \) so \(f\) has a maximum in that range and we can find it using simple calculus.
If the derivative is 0, then \( n p^{n-1} – (n+1) p^{n} = 0 \) so \( p = \frac n {(n+1)} \) at which point
\( f(p) = f(\frac n {n+1}) = (\frac n {n+1})^n (1 – (\frac n {n+1}))\) and we want this to be \( < \frac 1 n \).
Multiplying by n and rearranging we want
$$ n \frac {n^n (n+1) – n^{n+1}} {(n+1)^{n+1}} < 1 $$
or
$$ \frac{n^{n+2} + n^{n+1} – n^{n+2}} {(n+1)^{n+1} } < 1 $$
or
$$ \frac {n^{n+1}}{(n+1)^{n+1} }< 1 $$
which is self-evident for \(n\) an integer.
In fact, we can go further, just for fun, and note that
$$ \lim_{n \to \infty} { (\frac {n-1} n)^n} = \frac 1 e $$
