I The inequality in the Heisenberg uncertainty relation

[Derek P]

I was musing about why the HUP is an inequality. If you analyse a wave packet the spatial frequency spectral width is inversely proportional to the spatial width. So there should be an equality such as Heisenberg's equation 3 in this paper. Has anyone got a simple explanation of where the inequality comes from? I initially thought it was the fact that we square everything to get probability distributions, but I seem to have a mental block so any help would be appreciated.

Should this be a B level question?

 

[PeroK]

Any state has a variance for each observable. So, there is an equality there which you can calculate for any specific state.

The HUP, however, specifies a minimum across all states. Some states, e.g. Coherent states of the harmonic oscillator have an uncertainty equal to the minimum. Most states have an uncertainty greater than this.

 

[Derek P]

I was musing about why the HUP is an inequality. If you analyse a wave packet the spatial frequency spectral width is inversely proportional to the spatial width. So there should be an equality such as Heisenberg's equation 3 in this paper. Has anyone got a simple explanation of where the inequality comes from? I initially thought it was the fact that we square everything to get probability distributions, but I seem to have a mental block so any help would be appreciated.
Should this be a B level question?

One Indian meal later and mental block has gone. Forget it, thanks.

 

[vanhees71]

You can also calculate the wave packets, for which the equality sign holds. That's a way to introduce coherent and squeezed states!

 

[Derek P]

You can also calculate the wave packets, for which the equality sign holds. That's a way to introduce coherent and squeezed states!

Well, yes, once you have the infrastructure of Fock spaces in place :)

 

[vanhees71]

You don't need Fock spaces here. Just Schrödinger wave-mechanics is sufficient.

 

[Derek P]

You don't need Fock spaces here. Just Schrödinger wave-mechanics is sufficient.

Yes, you're right.