# Why does a Gaussian wave packet take on the minimum value of the Heisenberg Uncertainty Principle?

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Here we have:

#psi(x) = 1/(2pisigma^2)^"1/4" e^(-(x-x_0)^2/(4sigma^2) + i/ℏ p_0(x-x_0))#

which is normalized over allspace:

#int_(-oo)^(oo) psi^"*"(x)psi(x)dx = 1#

The Heisenberg Uncertainty Principle for momentum #p_x# and position #x# is:

#DeltaxDeltap_x >= 1/2 |i int_(-oo)^(oo) psi^"*"(x)[hatx, hatp_x]psi(x)dx|#

#>= ℏ/2#

where:

#[hatx, hatp_x]psi = -iℏx(delpsi)/(delx) + iℏ(del)/(delx)(xcdotpsi) = iℏpsi# , and #ℏ = h/(2pi)# .

Why do Gaussian wave packets take on the value #ℏ/2# exactly, not greater than it?

Here we have:

#psi(x) = 1/(2pisigma^2)^"1/4" e^(-(x-x_0)^2/(4sigma^2) + i/ℏ p_0(x-x_0))# which is normalized over allspace:

#int_(-oo)^(oo) psi^"*"(x)psi(x)dx = 1#

The Heisenberg Uncertainty Principle for momentum

#DeltaxDeltap_x >= 1/2 |i int_(-oo)^(oo) psi^"*"(x)[hatx, hatp_x]psi(x)dx|#

#>= ℏ/2# where:

#[hatx, hatp_x]psi = -iℏx(delpsi)/(delx) + iℏ(del)/(delx)(xcdotpsi) = iℏpsi# , and#ℏ = h/(2pi)# .

Why do Gaussian wave packets take on the value

##### 2 Answers

See link . It offers a very nice derivation of this fact.

#### Explanation:

Without repeating all of the mathematics contained in the above link, here is a very concise, higher level conceptual view:

When considering the Gaussian wave packet, we are considering a probability distribution of finding the particle. Initially (equivalent to saying

Hopefully these two sources offer you some guidance on this topic!

That a Gaussian wavefunction satisfies the minimum uncertainty bound can be verified by a direct calculation. What is perhaps more interesting is the question whether it is the only one to do so!

#### Explanation:

To understand this, we should go back to the derivation of the uncertainty principle itself. While we can do this for the specific position - momentum uncertainty principle that is being talked about here, it is no more difficult (and perhaps more instructive) to derive the general version.

**The general uncertainty principle**

To this end, let us start with any two Hermitean operators

Now, consider a function of the real number

Now,

so that even its minimum value must be non-negative. Thus we have

which can be immediately rewritten as

If we now replace

For the special case

the well known **position-momentum uncertainty relation**.

**Attaining the minimum uncertainty bound**

Let us now look for the condition that will allow a wavefunction to satisfy the minimum uncertainty bound, i.e.

For this, two things will have to happen

and

where

For the position momentum uncertainty relation, these become

and (in the position representation where

Defining

which leads to

Which is easily seen to be satisfied by

Demanding that the wavefunction be normalized leads to the condition :

so that the normalized wavefunction is given by

So, the upshot is - not only does the Gaussian wavefunction satisfy the minimum **must be Gaussian.**

**Time evolution and minimum uncertainty**

The next interesting question is "if a wavefunction leads to a minimum uncertainity state at a given instant, does this property persist in time?"

The answer depends, of course, on the Hamiltonian. After all, it is the Hamiltonian that, via the Schroedinger equation

dictates how a given initial state evolves in time.

It turns out that while for a free particle, an initial Gaussian wave packet evolves into another Gaussian one - but one for which

On the other hand, for a harmonic oscillator, and initially Gaussian wavefunction, may stay a Gaussian wavefunction with time dependent