On Heisenberg's uncertainty principle

  • I
  • Thread starter Ssnow
  • Start date
  • #1
Ssnow
Gold Member
509
149

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,289
5,104
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.
 
  • #3
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,376
5,969
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).
 
  • #4
jfizzix
Science Advisor
Insights Author
Gold Member
753
346
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.
 
  • #5
PeterDonis
Mentor
Insights Author
2019 Award
28,239
8,026
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.)
 
  • #6
jfizzix
Science Advisor
Insights Author
Gold Member
753
346
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:
[itex]\rho (x,x') = \langle x|\psi\rangle\langle \psi|x'\rangle = \psi(x)\psi^{*}(x') [/itex]
The example that comes to my head is the statistics of a correlated double-gaussian wavefunction in [itex]x[/itex] and [itex]y[/itex], when you trace over [itex]y[/itex]:
[itex]\rho (x,x') = \int dy\; \psi(x,y)\psi^{*}(x',y) [/itex]
Since we contrive it to be correlated, the joint wavefunction doesn't factor, i.e., [itex]\psi(x,y)\neq\psi(x)\psi(y)[/itex]
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)
 
  • #7
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
14,376
5,969
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.
 

Related Threads on On Heisenberg's uncertainty principle

Replies
14
Views
2K
Replies
8
Views
6K
Replies
3
Views
1K
  • Last Post
Replies
22
Views
6K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
2
Replies
38
Views
7K
  • Last Post
Replies
22
Views
5K
  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
3
Views
708
Top