Computing the expectation value of momentum, kinetic energy, and compute

Hi laser123,

Recall that the definition of the expectation value [itex] \langle A \rangle [/itex] for any operator A is defined as [itex] \langle A \rangle = \langle \psi | A | \psi \rangle [/itex]. Do you know the operator forms of momentum, position, and kinetic energy in the position basis?

Additionally, recall that [itex] \Delta A = \sqrt{ \langle A^2 \rangle - \langle A \rangle^2 }[/itex]. Once you calculate the square of the expectation value and the expectation value of the square of the position and momentum, you can do some simple calculations to solve for [itex] \Delta x \Delta p[/itex].

Hope this helps!

 

Remember that, for the transition from classical dynamical variables to quantum dynamical variables, we replace the respective variable with its operator (where applicable).

In the X basis, when we act on a ket [itex] | \psi \rangle [/itex] with the position operator, we obtain the position. As an eigenvalue equation, [itex] X | \psi \rangle = x | \psi \rangle [/itex].

The classical representation of kinetic energy is [tex] KE = \frac{1}{2} m v^2 = \frac{1}{2} \frac{p^2}{2m} [/tex] The transition to quantum mechanics, then, requires inserting the momentum operator [itex] \hat{p} = -i \hbar \frac{\partial}{\partial x}[/itex]. (This is for the 1-D momentum operator in the position basis.) The kinetic energy then becomes [tex] KE = \frac{1}{2} \frac{\hat{p}^2}{2m} = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} [/tex]

Now simply use the expectation value equations in order to solve for the expectation values of momentum and kinetic energy.

 

Before we do the dummy walkthrough, can you show me how you computed the expectation values for the momentum and kinetic energy? Since this wavefunction is a little weird, I'm wondering if I'm calculating it correctly myself.

 

Sorry -- is the wavefunction supposed to be [itex] \psi(x) = \sqrt{\frac{30}{L^5}} x (L-x) [/itex]? From the calculation you just did, it looks like that. But your original post is [itex] \psi(x) = \sqrt{\frac{30}{L^5}} (L-x) [/itex]. Frankly, I like the former more than the latter.

 

The forum rules require you show some effort at attempting the problem yourself. Jmceive's already given you enough information in this thread for you to make some attempt at solving the last part of the problem.