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(adsbygoogle = window.adsbygoogle || []).push({});

\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?

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# Boson statistics and the uncertainty principle

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