Finding the State and Expectation Value for a Free Particle at Time t

[spacetimedude]

Homework Statement

The eigenstates of the momentum operator with eigenvalue k are denoted by |k>, and the state of the system at t = 0 is given by the vector
$$|{ψ}>=\int \frac {dk}{2π} g(k)|{k}>$$
Find the state of the system at t, |ψ(t)>.

Compute the expectation value of $\hat{P}$.

Homework Equations

The Attempt at a Solution

From what I learned from the lecture, I just have to introduce (multiply) $\exp[\frac{-i}{\hbar}\hat{H}t]$ where in this free particle case, $\hat{H}=\frac{\hat{P}^2}{2m}$, to |ψ>.

So $$|{ψ(t)}>=\exp[\frac{-i}{\hbar}\frac{\hat{P}^2}{2m}t]\int \frac {dk}{2π} g(k)|{k}>$$

When I compute for the expectation value using $<ψ(t)|\hat{P}|ψ(t)>$, I get $\frac{1}{4\pi^2}\int |k|^2 \hat{P} dx$.

The exponentials cancel due to multiplying of its complex conjugate.
I was confused on how to get rid of the two integrals with dk. I assumed (without reason so probably wrong) they become 1 because they are the product of complex conjugate and the total probability is 1.

Any help will be appreciated!

PS. How do I type ket in latex?

 

[BvU]

Hi,

You want to be a bit more accurate.

For time development I think you want $H$ and not $\hat H$. $H$ is an operator and yes, the $|k>$ are simultaneous eigenfunctions of H and p . See the link for the normalization: you don't have two integrals with $k$ but one with $k$ and one with $k'$.

Equally, for $\hat P$ you want the operator $\ p\$ and not the number $\hat P$ (that would be kind of circular ...)

Ket in $LaTeX$ is just what you did. Or use \left \langle a\middle | b\right \rangle : $\ \ \left \langle a\middle | b\right \rangle$

 

[spacetimedude]

Hi,

You want to be a bit more accurate.

For time development I think you want $H$ and not $\hat H$. $H$ is an operator and yes, the $|k>$ are simultaneous eigenfunctions of H and p . See the link for the normalization: you don't have two integrals with $k$ but one with $k$ and one with $k'$.

Equally, for $\hat P$ you want the operator $\ p\$ and not the number $\hat P$ (that would be kind of circular ...)

Ket in $LaTeX$ is just what you did. Or use \left \langle a\middle | b\right \rangle : $\ \ \left \langle a\middle | b\right \rangle$

Hmm, I'm a bit confused. From my understanding, $\hat{H}=\hat{T}+\hat{U}$ where $\hat{T}$ is the kinetic operator and $\hat{U}$ the potential operator. And for free particle, U=0. Then I wrote the kinetic energy operator in terms of the momentum operator.
Isn't $\hat P$ the operator $-i \hbar \frac {d}{dx}$ and not a number? And do you mean that when I take the complex conjugate of one of the $Ψ(t)$, I should take the integral of k' and the other one in k, resulting in $\frac{1}{4\pi^2}\int dk' g(k') \int dk g(k)$ ? I don't know why I need to consider $\hat{p}^2$ because it gets canceled out when multiplying the complex conjugate of the exponential.

 

[BvU]

Sorry, my bad -- I wrongly interpreted your notation . So $\hat P|k> = k |k>$ . That way you can describe the time development of $|k>$.
For the time development, the exponent therefore stays under the integral: each $|k>$ has its own time development.

And do you mean that when I take the complex conjugate of one of the Ψ(t)Ψ(t)Ψ(t), I should take the integral of k' and the other one in k, resulting in

yes. see the link: the integration yields something with a delta function.

 

[spacetimedude]

Sorry, my bad -- I wrongly interpreted your notation . So $\hat P|k> = k |k>$ . That way you can describe the time development of $|k>$.
For the time development, the exponent therefore stays under the integral: each $|k>$ has its own time development.

yes. see the link: the integration yields something with a delta function.

Not quite sure how to follow the steps on the link. The integral I have for the expectation is $\frac {1}{(2\pi)^2}\int dx \int dk \int g(k')g(k)k'k(dk')$. Have I done something wrong?

 

[BvU]

Where are the $\ \ \left \langle k'\middle | k \right \rangle \ \$? $\quad$ And the operator only works to the right, doesn't it ?

 

[spacetimedude]

Where are the $\\\ \left \langle k'\middle | k \right \rangle \ \$? $\quad$ And the operator only works to the right, doesn't it ?

Hmm, is it supposed to be $<k'| \frac{1}{2\pi} \int dk' g(k') \exp{\frac{i\hat P ^2 t}{2m\hbar}} \int \hat P \frac{1}{2\pi}dk g(k) \exp{\frac{-i\hat P^2t}{2m\hbar}}|k>$ then the exponentials cancel so $<k'| (\frac{1}{2\pi})^2 \int dk' g(k') \ \int \hat P dk g(k)|k>$? Do I then compute for the expected value using the method of integrating the products of the complex conjugate of <k'|..., $\hat P$ and ...|k>?

Becomes $\frac{1}{4\pi^2}\int dx k'^* [\int \int dk'dk \hat P g(k')g(k)k]$

 

[BvU]

I also wonder how you can end up with a triple integral
Did you find the normalization eqn for the momentum eigenstates in the link ?

 

[BvU]

Note that for $<p>$ (expectation value for $\hat p$ ), you want to end up somewhere at $\ \int k\;|g(k)| ^2\;dk\$ (according to my Merzbacher, QM, $\ |g(k)| ^2$ is the probability density in momentum space)