The Momentum Operator

Thursday, November 29, 2018

The Momentum Operator

Momentum

What if we think of a new linear operator corresponding to ∂/∂x (differentiating the wave function with respect to position).
This does not correspond to an observable because it is anti-hermitian.
(If an operator is anti-hermitian, <A|I|B> = -<B|I|A>*)
When you x by -i however, the operator becomes hermitian.
This means -i∂/∂x is an hermitian operator and so must correspond to some observable.

Let's verify this to be sure
<ψ|P|ψ> is always real

= ∫ψ* -i∂/∂x ψ

= -i∫∂ψ/∂x ψ*

If this is always real, it must be its own complex conjugate.
This is the case as it's complex conjugate is
i∫∂ψ*/∂x ψ

So now we want to know the eigenvectors of -i∂/∂x to help us figure out what observable -i∂/∂x corresponds to.

-i∂ψ/∂x = kψ

Solving this equation gives
ψ = e^ikx.
ψ = cos(kx) +isin(kx)

This function is an eigenvector of the operator -i∂/∂x and so corresponds to a particle being in a particular state.

The wavelength of the oscillations is the distance before cosx does a cycle. You have to move 2pi for one cycle and so

kλ = 2pi

λ=2pi/k

De broglie also gave us the equation p =h/λ

And so combining these gives

p = ħk

where ħ is the reduced Planck constant (h/2pi)

K is therefore the momentum of a particle in units of ħ.

We can write our original equation as

-iħ∂ψ/∂x = pψ

ψ = e^ipx/ħ.

Let's now consider a particle on a circle. For any state vector, the inner product on itself must be one. Let's therefore take the inner product on a state vector for a particle on a circle.

∫ψψ*dx

ψψ* is 1 and ∫dx is 2pi r.
This means we have to divide the wave function by √2pi r to ensure the inner product is 1. This wave the sum of the probabilities is always 1.

If we have a particle on a circle we want a periodic wave function as well. So let's look at what constraints have to be put on the wave function for it to be periodic.

e^ipx/ħ/√2pi r = e^{ip(x + 2pi r)/ħ}/√2pi r

e^ipx/ħ = e^{ip(x + 2pi r)/ħ}
e^{ip 2pi r/ħ} = 1

e^{ip 2pi n} = 1

and n is an integer so the general solution for a periodic wave function is

pr / ħ = n

p = ħn/r

Therefore the possible allowable values for p come in integer multiples of ħ/r. If any other values for p are chosen, the wave function won't be periodic and so won't correspond to motion on a circle. This tells us that momentum is actually a discrete quantity.
Quantum mechanics quantised waves and tells us that they actually come in discrete quanta. Now we have gone one step further to say the momentum of an object also comes in discrete chunks.

This also applies to straight lines because despite them not having radii, we can think of them as circles with increasing radii. This means if we have a particle moving along a straight line, the momentum appears more and more continuous For an infinitely long straight line (a circle with infinite radius), the particle would have perfectly continuous momentum.

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