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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?

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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

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  • $\begingroup$ 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 Apr 19 '17 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 ? $\endgroup$ – FireFistAce Apr 19 '17 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 Apr 20 '17 at 2:03
  • $\begingroup$ @PCat27 do you mean why we don't explicitly take d/dt(px) or d/dt(x) in the integral ? $\endgroup$ – FireFistAce Apr 20 '17 at 2:31
  • $\begingroup$ Yes, sorry my explanation. $\endgroup$ – PCat27 Apr 20 '17 at 2:42

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