Question

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

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?

Answer 1

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 `t = 0`), the particle is localized in such a way that the gaussian distribution takes on the minimal value of the uncertainty principle. However, at later times the probability distribution undergoes a "spreading", which I will offer this link to explain. The Gaussian wave packet will not always take on the minimum value of the uncertainty principle. This is only the case when the Gaussian wave packet is initialized.

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

Answer 2

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 `hat A` and `hat B`. Since their commutator is anti-Hermitean, we can define another Hermitean operator by

`[hat A, hat B] =i hat C`

Now, consider a function of the real number `lambda` defined by the expectation value

`f(lambda) = <(hat A-i lambda hatB) (hatA+i lambda hatB) >`
`qquad = < hatA^2 >+lambda^2< hatB^2 >+i lambda < hatAhatB-hatBhatA>`
`qquad =< hatA^2 >+lambda^2< hatB^2 > - lambda < hatC >`
`qquad = < hatA^2 > +< hatB^2 >(lambda^2-lambda {< hat C >}/{< hatB^2 >})`
`qquad = < hatA^2 > -{< hat C >^2}/{4< hat B^2 >}+< hatB^2 >(lambda^2-lambda {< hat C >}/{< hatB^2 >}+{< hat C >^2}/{4(< hat B^2 >)^2})`
`qquad = < hatA^2 > -{< hat C >^2}/{4< hat B^2 >} + < hatB^2 >(lambda -{< hat C >}/{2< hatB^2 >})^2`
`qquad >= < hatA^2 > -{< hat C >^2}/{4< hat B^2 >}`

Now, `f(lamda)>=0`, since it is nothing but

`f(lambda) = int_-oo^oo psi^"*" (hat A-i lambda hatB) (hatA+i lambda hatB)psi dx`
`qquad int_-oo^oo | (hatA+i lambda hatB)psi|^2 dx`

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

`< hatA^2 > -{< hat C >^2}/{4< hat B^2 >} >= 0`

which can be immediately rewritten as

`< hatA^2 >< hatB^2 > >= {< hat C >^2}/4`

If we now replace `hatA` and `hatB` by `hatA - < hatA >` and `hatB -< hatB>`, respectively (it is easy to see that this does not change the commutator), then we end up with the generalized uncertainty principle

`color(red)(Delta A Delta B >= | < hatC >|/2 )`

For the special case `hatA = hatx` and `hatB =hatp`, we have `hat C = -i[hatx,hatp] = ℏ` and thus

`color(blue)(Delta x Delta p >= ℏ/2)`

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.

`Delta A Delta B = | < hatC >|/2`

For this, two things will have to happen

`f(lambda)=0 implies (hatA - < hat A>+i lambda (hatB- < hatB >))psi = 0`

and

`lambda -{< hat C >}/{2(DeltaB)^2}=0 implies lambda = {DeltaA}/{DeltaB}times "sgn"(< hatC >)`

where `"sgn"(x)` stands for the sign of `x`.

For the position momentum uncertainty relation, these become

`lambda = {Delta x}/{Delta p}`

and (in the position representation where `hat x = x` and `hat p = ℏ/i del/{del x}`, and using the notation `x_0 = < hat x >, p_0= < hat p >`)
`(x-x_0 + i {Delta x}/{Delta p} (ℏ/i del/{del x} -p_0)) psi(x,t)= 0`

Defining `{Delta x}/{Delta p} ℏ = 2 sigma^2`, we rewrite this equation as

`((x-x_0)/(2 sigma^2)-ip_0/ ℏ + del/{del x}) psi(x,t) = 0`

which leads to

`(d psi)/psi = - ((x-x_0)/(2 sigma^2)-ip_0/ ℏ )dx`

Which is easily seen to be satisfied by

`psi(x,t) = A(t) exp(- (x-x_0)^2/(4sigma^2)+ip_0/ ℏ x)`

Demanding that the wavefunction be normalized leads to the condition :

`|A(t)|^2 = 1/sqrt{2pi sigma^2}`

so that the normalized wavefunction is given by

`psi(x,t) = 1/root{4}{2pi sigma^2} exp(- (x-x_0)^2/(4sigma^2)+ip_0/ ℏ x)`

So, the upshot is - not only does the Gaussian wavefunction satisfy the minimum `x`-`p` uncertainty bound as long as `sigma` is real, any wavefunction that satisfies the minimum `x`-`p` uncertainty bound 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

`iℏ {del psi}/del t = hat H psi`

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 `sigma^2` is replaced by a complex quantity. Thus, an initial minimum-energy wavepacket evolves into a state which no longer gives minimum uncertainty product.

On the other hand, for a harmonic oscillator, and initially Gaussian wavefunction, may stay a Gaussian wavefunction with time dependent `x_0` and `p_0` values. The simplest case of course is the ground state energy eigenfunction for the harmonic oscillator (for this, of course, `x_0` and `p_0` both remain fixed at zero). However, you can have more general cases - the so called coherent and squeezed coherent states of the harmonic oscillator which maintain the minimum uncertainty product as time evolves.