# Minimum uncertainty state?

1. May 21, 2007

### pivoxa15

What is a minimum state in QM? A state being a state vector. I know observables can have an average uncertainty value attached to it but a whole state?

What would it mean in QM for a state to achieve minimum uncertainty?

2. May 22, 2007

### Fra

If your thinking of minimum uncertainty like minimum entropy? then any "pure state" has entropy zero.

It may still be a little relative since the von neumann entropy is a measure of uncertainty (implicitly) relative to the chosen partitioning of the space of distributions so to speak, where the partitions are assume a priori equally likely. So you would need to specify asking for a minimum entropy relative a given equiprobable partitioning.

If we toy with the idea that for any given state, one can find a partition where this state is a pure state, then one can get zero entropy by choice of partition. So the von neumann is relative to a choice (of yours) of partitioning of distribution space where you prior info is assumed to be uniformly distributed.

/Fredrik

3. May 22, 2007

### pivoxa15

I am not thinking about entropy. I think I am referring to when the wavefunction is such that del(x)del(p)=(hbar)/2

4. May 22, 2007

### Fra

I see. That should be for a gaussian probability distribution. The fourier transform of a gaussian distribution is still gaussian. So a gaussian x-distribution induces a gaussian p-distribution, and their standard deviations satisfy the minimum uncertainty.

/Fredrik

5. May 22, 2007

### pivoxa15

Is that if and only if condition, gaussian distribution <=> minimum uncertainty state? What does it mean physically for a state to have minimum uncertainty?

6. May 22, 2007

### Mentz114

In quantum optics, states of minimum uncertainty are call 'squeezed' states. They can be created in practice using PDC. I'd tell you more if I knew more, but I haven't got that far in QO.

Last edited: May 23, 2007
7. May 23, 2007

### Fra

Yes the gaussian function family is the only function satisfying this.

I guess you can elaborate "the meaning" that in different ways. One can note that the gaussian function is famous for it's properties and connections to statistics (normal distributions and central limit theorem) and probability theory as beeing a self-conjugate prior, making it special.

/Fredrik

8. May 23, 2007

### Anonym

The correspondent classical physical system (not the statistical ensembles) obey delta x*delta p = 0. Therefore the minimum uncertainty wave packets are the quantum objects most close to their classical counterparts. The transition is discontinuous and spontaneous (instant). It is known as the collapse of the wave packet.

Another remarkable property is that they are stable localized objects. Notice that it is quite general physical property: proton is stable, hydrogen atom is stable and isolated classical material point is stable.

Regards, Dany.

9. May 23, 2007

### pivoxa15

If a quantum state that evolves in time and space obeys being a minimum uncertainty state what would it mean physically? Or what would be a physical intepretation of this quantum state?

10. May 23, 2007

### Anonym

If you ask what is the physical state, I suggest reading E. Schrödinger, “THE PRESENT SITUATION IN QUANTUM MECHANICS” (1935) in W&Z, p.152.

I have no idea about its physical interpretation, but if you want to see it, look to the micrographs (b),(c),(d) Fig.3 in the results presented by A. Tonomura et al. “Double-biprism electron interferometry”, App. Phys. Lett., 84(17), 3229 (2004).

Regards, Dany.

11. May 24, 2007

### pivoxa15

How sure are you on this? You could have a state that is not a gaussian but modulated by a gaussian function. i.e a gaussian times another function. The product is not a gaussian but its amplitude is modulated by a gaussian like excited eigenstates of the harmonic oscillator.

Last edited: May 24, 2007
12. May 24, 2007

### marlon

I don't know, never heard of it. You might wanna provide some additional info to clarify the context.

marlon

13. May 24, 2007

### Fra

There are probably several proofs of this.

First one need to note that specific meaning of the "delta x" uncertainty in HUP is the standard deviation of x and p respectively.

One trick is to define an operator

$$g = (x-a)/2k + k(\frac{\partial}{\partial x}-c)$$ a,k,c constants.

Then of course the expectation $$<gg\dagger> >= 0$$

Then you can see that the only nonzero state yielding zero exepectation is the gaussian.

Then you can relate this to the product of the standard deviations of x and p, by choosing the constants a,k,c. a = <x>, c=i<p>/hbar, k = standard deviaton of x or something like that.

A lineout of this is at http://arxiv.org/PS_cache/quant-ph/pdf/0301/0301057v1.pdf

/Fredrik

Last edited: May 24, 2007
14. May 24, 2007

### jambaugh

Every quantum mode represented by a wave-function or "state"-vector is minimally uncertain i.e. maximally specified. We call this a sharp mode.
In order to express less than maximal modes i.e. less than minimal uncertainty one must use a density operator.

For a sharp mode: $$\psi$$
The corresponding density operator is: $$\rho=\psi\otimes\psi^\dag$$
More generally the density operator is a weighted sum of such sharp operators. These weights are "classical probabilities".

There is then an entropy (Von Neumann entropy) which is a good measurement of the level of uncertainty above minimal:
$$E_{VN} = -k_B \mathop{Tr}(\rho \ln \rho )$$
It is indeed zero for the sharp modes.

Now I think you are thinking of uncertainties in terms of some specific observables namely p and x. Keep in mind that every mode ("state") minimizes the uncertainty of some set of observables.

If you are thinking in terms of some specific observable then the eigen-modes minimize the uncertainty of that observable. If you are thinking of some pair of observables then the mode with with minimum magnitude of its expectation value for the commutator of those observables will minimize the product of uncertainties in the two observables.

If you are thinking in some other terms you'll have to get specific in what you are asking.

Regards,
James Baugh

15. May 24, 2007

### StatMechGuy

As has been mentioned, "squeezed states" in quantum optics are an example of a minimum uncertainty state. Coherent states are a special case of squeezed state, where the uncertainty is equally distributed between the canonically conjugate coordinates.

It's worth noting that this can refer to number/phase uncertainty as well as position/momentum uncertainty.

I believe the definition of a minimum uncertainty state is one in which the Heisenberg uncertainty relation is an equality, not an inequality.

16. May 25, 2007

### pivoxa15

I want to get some terminology correct.

Minimum uncertainty state: del(p)del(x)=hbar/2

coherent state: del(p)=del(x)=sqrt(hbar/2) which are rare cases?

squeezed coherent state: del(p) and del(x) can be arbitarily small or large but maintaining a constant product hence still a minimum uncertainty state. More common?

Are the above definitions correct?

17. May 25, 2007

### Anonym

I doubt that it is correct. Phase is not observable and can’t be the canonically conjugated to number. I believe that you should consider number/phase difference uncertainty (ties?). However, I understand close to nothing in non-self ajoint operators.

Regards, Dany.

18. May 25, 2007

### Mentz114

Anonym said :
Anonym, phase and photon number are conjugate operators. This is a basic result in quantum optics. The number eigenstates form a complete basis for coherent states.

19. May 25, 2007

### Anonym

I will appreciate it greatly if you will provide the reference for that basic result in quantum optics. I will appreciate it also if it will turns out that the result is far from being basic and even far from being result.

To the best of my knowledge we discuss the open problem which I consider the most interesting in QO and use as a laboratory to understand physics described by non-self adjoint operators. Mainly, I refer to M&W (Ch. 21 in particular) and P. Carruthers, M. Nieto, Rev. Mod. Phys.,40, 411 (1968) (paragraph 5 “Operators of number and phase for the harmonic oscillator).

Regards, Dany.

20. May 25, 2007

### Mentz114

Dany, I'm sorry if the term 'basic' has offended your sensibilities. Please look in 'Introductory Quantum Optics', Gerry and Knight, CUP 2005, Chapter 2 'The Quantum Phase'.

You will see there that you are right. The apparent conjugacy of the number and phase leads to a contradiction because of the non-hermiticity of the operators. As I remarked earlier, I'm a beginner on QO, while you obviously are not.