# $\langle xp+px\rangle|_{t=0}=2\langle p\rangle\langle x\rangle|_{t=0}$ for the free particle?

Quantum Mechanics, Volume 1 by Claude Cohen-Tannoudji, Bernard Diu and Frank Laloe. Complement L-III, exercise 4 (page 342).

Basically consider a free particle, and calculate the variance(uncertainty) of $$(\Delta X)^2$$ by using Ehrenfest's theorem.

I got $$\langle X\rangle =\frac{\langle p\rangle }{m}t+K, \langle X^2\rangle =\frac{\langle p^2\rangle }{m^2}t^2+\frac{c_1}{m}t+c_2$$ and $$\langle xp+px\rangle =\frac{2}{m}\langle p^2\rangle t+c_1 ,$$where $$c_1,c_2,K$$ are arbitrary constant from integration with respect to time $$t$$.

The answer in the text book is $$(\Delta X)^2 =\frac{1}{m}(\Delta p)_0^2t^2+(\Delta X)_0^2.$$

According to my calculation, there is an extra term $$(\frac{c_1}{m}-\frac{2\langle p\rangle }{m}K)t .$$

If I assume $$c_1=K=0$$, then it's done.

However, this does not seem to be the case the textbook suggested, instead, it seemed to be suggesting that $$\langle xp+px\rangle |_0=c_1=2\langle p\rangle K=2\langle p\rangle \langle x\rangle |_0 .$$

(Notice $$\langle p\rangle$$ and $$\langle p^2\rangle$$ are independent of time. Further, if one assumes transitional symmetry, then as a term in measurement $$(\frac{c_1}{m}-\frac{2\langle p\rangle }{m}K)t$$ must disappear, so even without the context of the textbook $$\langle xp+px\rangle |_0=c_1=2\langle p\rangle K=2\langle p\rangle \langle x\rangle |_0$$ for a free particle.)

My question is:

1. How to prove $$\langle xp+px\rangle |_0= 2\langle p\rangle \langle x\rangle |_0$$

2. If the expression is not true, what happened to the transitional symmetry?

(* the subscript $$0$$ meant measure at time $$t=0$$)

• @CosmasZachos for free particle assume $V=constant=0$ for simplification. The homework point out that the explicit time dependency essentially reduced to the explicit time dependence of $<p>$ and $<p^2>$ after taking $\frac{d}{dt}<p>$ and $\frac{d}{dt}<p^2>$, both of them equatl to a constant. Then one may try to integrate back. (Just decompose it a little further and then one may integrate the functions as polynomials.) – J C Sep 28 '18 at 14:09
• It's OK. You did your QM right, but you dismissed the "with a suitable choice of the time origin" instruction. It is just a glorified change of variables reducing to their expression. You really want me to do it for you? – Cosmas Zachos Sep 28 '18 at 15:20
• @CosmasZachos I thought about that as well, but you see, in QM there is transitional symmetry. So the $K$ in $<x>$ can be changed arbitrarily, and it does not have to be $0$, but as one might have thought, moving the reference frame of calculation by $K$ would not change the $(\Delta X)^2$. Therefore, come back to the case, $<xp+px>|_0=2<x><p>|_0$ must hold, or the transitional symmetry is nolonger hold. If transitional symmetry is nolonger hold, there is a much bigger question... So... how to prove $<xp+px>|_0=2<x><p>|_0$ ? – J C Sep 28 '18 at 17:08
• No, you certainly do not wish to prove that. The book certainly does not suggest that in any way. See my answer. – Cosmas Zachos Sep 28 '18 at 18:48
• @CosmasZachos I'm not saying that your approach won't work to obtain the solution in the textbook. I'm saying that. (If) The textbook is correct, (then) and $<px+xp>|_0=2<p><x>|_0$. – J C Sep 28 '18 at 19:22

Your text instructs you to heed "with a suitable choice of the time origin". Certainly not to nullify the extra term $$A_0 t\equiv \frac{t}{m} \langle \{(p-\langle p\rangle_0),( x-\langle x\rangle_0) \}\rangle_0$$ you correctly found. Instead, to absorb it into a shifted time variable.
To wit, from your original $$(\Delta X)^2 _t=\frac{1}{m}(\Delta p)_0^2~t^2+(\Delta X)_0^2 +A_0 ~t,$$ pick a time $$\tau$$, so $$(\Delta X)^2 _\tau=\frac{1}{m}(\Delta p)_0^2~\tau^2+(\Delta X)_0^2 +A_0 ~\tau,$$ which you may subtract from the above to get $$(\Delta X)^2 _t -(\Delta X)^2 _\tau =\frac{1}{m}(\Delta p)_0^2~(t^2 -\tau^2)+A_0 ~(t-\tau)\\ = \frac{1}{m}(\Delta p)_0^2~(t -\tau ) ^2 + \left ( \frac{2\tau}{m}(\Delta p)_0^2 +A_0\right ) ~(t-\tau) .$$
You then solve for $$\tau$$ to nullify the big parenthesis, $$\tau =-mA_0/2 (\Delta p)_0^2 = \frac{\langle p\rangle \langle x\rangle_0 -\langle xp+px\rangle_0/2}{(\Delta p)^2_0} = \frac{-\langle \{(p-\langle p\rangle_0),( x-\langle x\rangle_0) \}\rangle_0}{2(\Delta p)^2_0},$$ substituting your initial conditions. Note this does not vanish for all distributions (initial conditions): it it absolutely not constrained by the TDSE, or measurements, or anything else. It is a constant.
You then see that for this origin of time, $$(\Delta X)^2 = (\Delta X)^2 _\tau + \frac{1}{m}(\Delta p)_\tau^2~(t -\tau ) ^2 ,$$ your book's expression, recalling that $$(\Delta p)_0=(\Delta p)_\tau$$.
• It doesn't matter what the book stated, we knew by changing $K$(make reference frame/ initial origion further) would not change the measurement $(\Delta X)^2$ by transitional symmetry. Thus the extra $t$ trem which dependent on the condition of $K$ must disappear. Change the time is esentially shift in intial position, but the things is that it should work for all intial position. Change $t$ didn't solve the question. I still put an upvote though. – J C Sep 28 '18 at 19:18
• Your answer would be changing if your $\tau$ is changing right? But that's not correct. Because if you repeated the experiement twice, with changing $\tau$, in my case $K$. $(\Delta X)^2$ should stay the same. The change of reference frame should not affect the experimental outcome. Otherwise physicsts in Europe would have a different science that of US. – J C Sep 28 '18 at 19:28
• Sigh... Yes, that's what "suitable" is all about. Only the $\tau$ solved for reduces the answer to that of the book. The authors know what they are asking. – Cosmas Zachos Sep 28 '18 at 19:32