# Is information lost in the Heisenberg's uncertainty?

If I set a particle at a known momentum and a known position, then I would expect that I know the precise position and momentum upon measurement (assuming I can measure that well). The uncertainty principle states that a known precise position means an unknown momentum. However, information cannot be created nor destroyed so somehow I am getting only 1 output from 2 inputs. This also means that this is an irreversible system. In a non-quantum example (so that I exclude the observer effect completely), I cannot know both the precise frequency and the precise time of a waveform.

If information isn't lost, could it perhaps be that I haven't given it that information in the first place, and that the information of the precise frequency and time can't exist together in the form of a wave?

Note: I am not a physics student, just a curious person. So if equations are being used, please try to explain what they represent. Thank you!

## 2 Answers

You can't "set" a particle at both a known momentum and a known position.

In your classical wave analogy, that would mean creating a wave at precise time and frequency. That's not a technical impossibility : it just doesn't make sense.

If the wave has unique time (think: a bang), it can't have a unique frequency ; to have a unique frequency (think: a well-defined tone) it must stay on for multiple periods, so its time can't be well defined. So, from a wave point of view, it can't be done.

No information is destroyed ; information just isn't there to begin with.

Generally, when performing "thought experiments", you have to take care of the implications of the setup. In this case, setting a particle with a known position and momentum mandates using some apparatus with a known position and momentum ; this can't exist per Heisenberg's uncertainty. That's the same catch with maxwell's daemon for example: when you realize that the daemon itself must stay at zero temperature, you understand why there is no paradox.

You cannot set a particle at a known momentum and a known position, that will vioate the uncertainty principle, there are no quantum states that satisfy your assumption. If you prepare a state with a known position, then the momentum of that state will be less defined, it corresponds to a state with multiple possible momentums, and viceversa.