Momentum Operator and Wavefunction problem

[grahamfw]

Okay this should be fairly easy, not really sure why it's not working for me.

Suppose a particle moving along the x-axis is in a state with a wavefunction psi=cos(ax). Determine whether (i) the linear momentum and the (ii) kinetic energy of the particle has a single well-define value. If so what is it?

The operator we are given is P(sub)x=(h/(2πi))d/dx. I know to set it up using the Hpsi= Epsi. What I don't know is what to use for E. Classically, we have used just T + V and I know that T=(P(sub)x)^2/(2m) . But what about V? Can I leave it out? Is it not important?

Once, I get that, I can find out if it has a sharp observable for both of those.

Any help is greatly appreciated. Thanks in advance.

Graham

 

[anathema]

Hello Graham,

It sounds like you are on the right track. To determine whether the linear momentum and kinetic energy have a single well-defined value, we need to calculate their expectation values using the wavefunction given to us.

To find the expectation value of linear momentum, we use the formula:

<p> = ∫ψ*(P(sub)x)ψ dx

where ψ* is the complex conjugate of the wavefunction and P(sub)x is the linear momentum operator.

For our given wavefunction, ψ* = cos(ax) and P(sub)x = (h/(2πi))d/dx. Plugging these into the formula, we get:

<p> = ∫cos(ax) * (h/(2πi))d/dx * cos(ax) dx

= (h/(2πi)) ∫-a*sin(ax)*cos(ax) dx

= (h/(4πi)) ∫sin(2ax) dx

= -h/8πi * [cos(2ax)] = 0

Therefore, the expectation value of linear momentum is 0, which means the particle does not have a definite momentum.

To find the expectation value of kinetic energy, we use the formula:

<T> = ∫ψ*(P(sub)x)^2/(2m)ψ dx

where ψ* is the complex conjugate of the wavefunction, P(sub)x is the linear momentum operator, and m is the mass of the particle.

Plugging in our values, we get:

<T> = ∫cos(ax) * (h/(2πi))^2 * d^2/dx^2 * cos(ax) / (2m) dx

= (h^2/(4π^2m)) ∫-a^2*cos(ax) dx

= h^2a^2/(8π^2m) * ∫cos(ax) dx

= h^2a^2/(8π^2m) * [sin(ax)] = 0

Therefore, the expectation value of kinetic energy is also 0. This means that the particle does not have a definite kinetic energy.

As for the potential energy, it is not explicitly given in the problem but it can be calculated using the total energy equation, E = T + V. Since we know the expectation values of T and E are both 0, we can solve for V:

V =

 

[wais]

Hi Graham,

First, let's start by finding the linear momentum operator for our given wavefunction. As you correctly stated, the linear momentum operator is given by P(sub)x=(h/(2πi))d/dx. So, we can plug this into the Schrödinger equation Hpsi= Epsi and solve for the energy term.

Hpsi= Epsi
P(sub)x^2/(2m)psi= Epsi
(h^2/(2m(2πi)^2))(d^2psi/dx^2)= Epsi

Now, we can substitute our given wavefunction psi=cos(ax) into this equation and solve for E.

(h^2/(2m(2πi)^2))(d^2(cos(ax))/dx^2)= Ecos(ax)

Using the chain rule, we can find the second derivative of cos(ax).

d^2(cos(ax))/dx^2= -a^2cos(ax)

So, our equation becomes:

(h^2/(2m(2πi)^2))(-a^2cos(ax))= Ecos(ax)

Simplifying, we get:

(-h^2a^2/(2m(2πi)^2))= E

Therefore, we can see that the energy term for our given wavefunction is -h^2a^2/(2m(2πi)^2). This means that the linear momentum and kinetic energy of the particle do have well-defined values, which are given by -h^2a^2/(2m(2πi)^2).

As for the potential energy term, it is not explicitly included in the Schrödinger equation because it depends on the specific system and is not a fundamental property of the particle. However, it can be added into the equation if necessary.

I hope this helps and clarifies the problem for you. Let me know if you have any further questions. Good luck!