On Heisenberg's uncertainty principle

[Ssnow]

Hi everybody, my question is a curiosity on the (generalized) Heisenberg principle:

$\sigma_{x}\sigma_{p} \geq \frac{\hbar}{2},$

where $x,p$ are the usual quantum operators and $\hbar$ the Planck constant divided by $2\pi$. If I understood correctly, Gaussian states that are solution of the quantum harmonic oscillator minimize the uncertainty principle with the equality. The question is what happen for the other Gaussian states? It is true for every Gaussian state? For example considering a Gaussian state of the form $\psi=Ce^{-\lambda x^2}$ (with $C$ constant and $\lambda$ a real parameter) it is the inequality true for every $\lambda$?
Thank you in advance for all answer to by doubt.

Ssnow

 

[PeroK]

Hi everybody, my question is a curiosity on the (generalized) Heisenberg principle:

$\sigma_{x}\sigma_{p} \geq \frac{\hbar}{2},$

where $x,p$ are the usual quantum operators and $\hbar$ the Planck constant divided by $2\pi$. If I understood correctly, Gaussian states that are solution of the quantum harmonic oscillator minimize the uncertainty principle with the equality. The question is what happen for the other Gaussian states? It is true for every Gaussian state? For example considering a Gaussian state of the form $\psi=Ce^{-\lambda x^2}$ (with $C$ constant and $\lambda$ a real parameter) it is the inequality true for every $\lambda$?
Thank you in advance for all answer to by doubt.

Ssnow

By varying the parameters of the oscillator you can get any positive value for $\lambda$.

In any case, it's straightforward to show that a Gaussian wave packet also exhibits the minimum uncertainty.

 

[vanhees71]

You can use any Gaussian for any system which admits position and momentum observables fulfilling the Heisenberg algebra, $[\hat{x},\hat{p}] =\mathrm{i} \hbar$.

For the harmonic oscillator they are special, because they are coherent (with $\Delta x \Delta p =\hbar/2$) or more general squeezed states (with the uncertainty oscillating between finite values).

 

[jfizzix]

If I understood correctly, Gaussian states that are solution of the quantum harmonic oscillator minimize the uncertainty principle with the equality. The question is what happen for the other Gaussian states? It is true for every Gaussian state?

For every pure Gaussian state it is true, but for Gaussian states made up of mixtures (not superpositions) of other wavefunctions, the uncertainty principle would not be saturated.

 

[PeterDonis]

Gaussian states made up of mixtures (not superpositions) of other wavefunctions

Can you give an example of such a Gaussian state? (I have no problem with states that are mixtures; I just don't see how the term "Gaussian" can be applied to them.)

 

[jfizzix]

Can you give an example of such a Gaussian state? (I have no problem with states that are mixtures; I just don't see how the term "Gaussian" can be applied to them.)

For a pure gaussian wavepacket, the density operator looks like:
\rho (x,x') = \langle x|\psi\rangle\langle \psi|x'\rangle = \psi(x)\psi^{*}(x')
The example that comes to my head is the statistics of a correlated double-gaussian wavefunction in x and y, when you trace over y:
\rho (x,x') = \int dy\; \psi(x,y)\psi^{*}(x',y)
Since we contrive it to be correlated, the joint wavefunction doesn't factor, i.e., \psi(x,y)\neq\psi(x)\psi(y)
The marginal position and momentum statistics are gaussian, but do not saturate the Heisenberg limit.

Correlated double-gaussian wavefunctions are commonly used in describing the position-momentum statistics of pairs of entangled particles. See for example: (https://arxiv.org/abs/1502.06996)

 

[vanhees71]

Hm, then you pose the interesting question, whether there are also non-pure states, i.e., proper mixed states with $\Delta x \Delta p=\hbar/2$ or whether there are only the pure states, which are described necessarily by Gaussian wave functions. I've to think about this.