0
$\begingroup$

I have two questions that i didn't find in books. When calculate time rates of change of the expectation values of $\langle x \rangle$ or $\langle p_x \rangle$, why is x or $p_x$ not derived from time?

Now i want to show this, if $\Psi(r, t) $ is a square integrable wave function normalised to unity,then:

$ \frac{d}{dt} \langle x² \rangle = \frac{1}{m} [ \langle xp_x \rangle + \langle p_xx \rangle]$

Using Ehrenfest's theorem:

$ \frac{d}{dt} \langle x² \rangle = \frac{d}{dt} \int \Psi^*(r,t)x^2 \Psi(r,t) dr $

Derive $\Psi^*(r,t)$ and $\Psi(r,t)$ from time and using schrödinger equation:

$ \frac{d}{dt} \langle x² \rangle = \frac{1}{2m i\hbar} \int \Psi^*[x^2 p_x² - p_x ² x^2 ]\Psi dr $

Inside the brackets is the operator [x²,p_x²], i don't know how calculate this operator, i believe that is not necesary, so:

$ \frac{d}{dt} \langle x² \rangle = \frac{1}{2m i\hbar} \int \Psi^*[x(xp_x)p_x - p_x(p_xx)x ]\Psi dr $

I can use $xp_x -p_xx= i \hbar$, but how if x and $p_x$ not commute?

$\endgroup$

1 Answer 1

1
$\begingroup$

Have you tried using commutator identities like $[A,BC] = [A,B]C + B[A,C]$ and $[AB,CD] = A[B,CD] + [A,CD]B = A[B,C]D + AC[B,D] + [A,C]DB + C[A,D]B $

Now in your case B = A and C =D

So the identity now reduced to $[AA,CC] = A[A,CC] + [A,CC]A = A[A,C]C + AC[A,C] + [A,C]CA + C[A,C]A $

with $x =A$ and $p_x = C $

$[x^2,p_x^2] = x[x,p_x^2] + [x,p_x^2]x = x[x,p_x]p_x + xp_x[x,p_x] + [x,p_x]p_xx + p_x[x,p_x]x $

using the identity $[x,p_x] = i\hbar$ we get $[x^2,p_x^2] = xi\hbar p_x + xp_xi\hbar + i\hbar p_xx + p_xi\hbar x $

Also note $[A,B] = -[B,A]$

References: https://en.wikipedia.org/wiki/Commutator

Also see: https://en.wikipedia.org/wiki/Ehrenfest_theorem

$\endgroup$
6
  • $\begingroup$ thanks!, i forgot completely that exist the identities of the commutators. but do you know why x or px is not derived in the expected value calculation? $\endgroup$
    – PCat27
    Commented Apr 19, 2017 at 20:19
  • $\begingroup$ @PCat27 I am not sure what you mean by "but do you know why x or px is not derived in the expected value calculation?" can you clarify that ? $\endgroup$ Commented Apr 19, 2017 at 20:25
  • $\begingroup$ When calculating the time derivate of the expectation values of ⟨x⟩ or ⟨px⟩, why is x or px not derived with respect to time? $\endgroup$
    – PCat27
    Commented Apr 20, 2017 at 2:03
  • $\begingroup$ @PCat27 do you mean why we don't explicitly take d/dt(px) or d/dt(x) in the integral ? $\endgroup$ Commented Apr 20, 2017 at 2:31
  • $\begingroup$ Yes, sorry my explanation. $\endgroup$
    – PCat27
    Commented Apr 20, 2017 at 2:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.