# Boson statistics and the uncertainty principle

1. Jul 24, 2013

### neerajareen

This question is regarding the boson statistics and it’s relation to the uncertainty principle. Consider we have a vacuum state and we apply a field operator on it to create a particle at position x, we end up with state like

\begin{array}{l}
\left| \psi \right\rangle = {\psi ^\dag }(x)\left| 0 \right\rangle \\
\left| \psi \right\rangle = \sum\limits_p^{} {{a^\dag }(p){e^{ - ipx}}} \left| 0 \right\rangle \\
\left| \psi \right\rangle = \sum\limits_p^{} {{e^{ - ipx}}} \left| p \right\rangle
\end{array}

So there is an equal probability for all momentum. And the result is completely consistent with the uncertainty principle.
Now consider a situation where we have created 99 bosons with momentum k. Using the statistics of creation of creation operators, when we now add an additional particle, we have

\begin{array}{l}
\left| \psi \right\rangle = {\psi ^\dag }(x)\left| {0,0,99,0,...} \right\rangle \\
\left| \psi \right\rangle = \sum\limits_p^{} {{a^\dag }(p){e^{ - ipx}}} \left| {0,0,99,0,...} \right\rangle \\
\left| \psi \right\rangle = {e^{ - i{p_1}x}}\left| {1,0,99,0,...} \right\rangle + {e^{ - i{p_2}x}}\left| {0,1,99,0,...} \right\rangle + {e^{ - i{p_k}x}}\sqrt {100} \left| {0,0,100,0,...} \right\rangle + {e^{ - i{p_{k + 1}}x}}\left| {0,0,99,1,...} \right\rangle
\end{array}/

We see that that when we create the now particle, it is more likely to have momentum k than any other momentum by a factor of 100. In the limit that we have a thousand billion particles in a certain momentum and then create a particle at a certain position, with almost 100% we can determine the momentum that it will have. Does this not violate the uncertainty principle?

2. Jul 24, 2013

### neerajareen

just to clarify

/dag is the dagger symbol (hermitian conjugate).

3. Jul 24, 2013

### stevendaryl

Staff Emeritus

What does your notation $\left| {0,0,99,1,...} \right\rangle$ mean?

4. Jul 24, 2013

### Cthugha

In order to have an effect of boson statistics, you need indistinguishable particles. So you do not get n individual particles, but one n-particle state. So in adding a particle, you go from the n-1 to the n-particle state. It is the position and momentum of the whole state of indistinguishable particles you need to consider. If this whole state has a pretty sharp momentum, it will be rather delocalized. If you (let us just assume that is possible somehow) create a particle that is localized better than the n-1 particle state, the added particle and the other particles cannot be completely indistinguishable, so the bosonic enhancement factor will not be n, but smaller.

5. Jul 29, 2013

### neerajareen

Thank you Cthuga. That makes sense