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In Sakurai's Modern Quantum Mechanics, p.270, he wrote an equation the parity transformation $\pi$ (where $\pi = \pi^\dagger = \pi^{-1}$) as $$\pi \left(1- \frac{i p \cdot d x'}{\hbar}\right) \pi^\dagger = \left(1+ \frac{i p \cdot d x'}{\hbar}\right). \tag{0}$$

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How is this consistent with these two equations (4.2.3) and (4.2.10) also derived:

$$ \pi x' \pi^\dagger = -x' \tag{1} $$

$$ \pi p \pi^\dagger = -p \tag{2} $$

Is that $$\pi dx' \pi^\dagger = dx' \tag{3} $$

$$\pi (p \cdot dx') \pi^\dagger = -( p \cdot dx' ) \tag{4} $$

How to understand the Eq. 1 versus Eq. 3? but how to understand the Eqs. 1, 2, versus Eq. 4?

Naively, it seems that

$$ \pi dx' \pi^\dagger = - dx' \tag{5},$$

because say $dx'=(x_A- x_B)$ is the spatial interval difference between two points on $A$ and $B$, then $$\pi dx' \pi^\dagger=\pi \Delta x \pi^\dagger=\pi (x_A- x_B) \pi^\dagger =(-x_A- (-x_B))=-(x_A- x_B)=-dx'.$$ Also I thought: $$ \pi (p \cdot dx') \pi^\dagger = +( p \cdot dx' ) \tag{6} $$

Could you correct me why Eqs. 3 and 4 are correct, but Eqs. 5 and 6 are not?

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  • $\begingroup$ For eq(`1) vs eq(3), the clue is in the statement: "$\textbf{p}$ is like mdx/dt, ... expect to be odd parity, like x'' $\endgroup$
    – paul230_x
    Commented Sep 12, 2021 at 15:56
  • $\begingroup$ that is NOT what I asked. look carefully. I am asking 𝜋(𝑝⋅𝑑𝑥′)𝜋†=−(𝑝⋅𝑑𝑥′) why not 𝜋(𝑝⋅𝑑𝑥′)𝜋†=+(𝑝⋅𝑑𝑥′)? $\endgroup$ Commented Sep 12, 2021 at 15:58
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    $\begingroup$ Now I see the possible answer: The $p$ is an operator in eq.4, while the $dx'$ is not an operator but just a number which is $\epsilon$ that can be treated as a number -- the $dx'=\epsilon$ does not flip under the parity transformation Thus eq3 and eq4 can be justified if what I said is true $\endgroup$ Commented Sep 12, 2021 at 16:11
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    $\begingroup$ Watch the caps! Eq(3) & (4) are correct. Simply use the definition of $T(dx)|x>=|x+dx>=(1-i\frac{p.dx}{\hbar})|x>$, and apply at eqn 4,2,8 and 4.2.9 . I just did, it worked $\endgroup$
    – paul230_x
    Commented Sep 12, 2021 at 16:17
  • $\begingroup$ your view is different than what I wrote in the earlier line. whether $p$ is an operator or it is a number here? $\endgroup$ Commented Sep 12, 2021 at 16:20

2 Answers 2

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With due respect to the formidable JunJohn S, his proof is doomed to unleash dyslexia demons of this kind, which I'd rather not deal with, and neither should you.

From his (4.2.5), $$ \pi | x\rangle= |-x\rangle, $$ and $$ \hat p = \int\!\! dx ~|x\rangle (-i\hbar\partial_x)\langle x| , $$ you immediately see $$ \pi \hat p \pi^\dagger = \int\!\! dx |-x\rangle (-i\hbar\partial_x)\langle -x|= \\ - \int\!\! dx |x\rangle (-i\hbar\partial_x)\langle x| =- \hat p, $$ by bland change of dummy variables of integration (and integration limits at infinity).

And that's that.

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Infinitesimal translation operator $\hat{T}(dx):=\left(1-i\frac{\hat{P}.\hat{dx}}{\hbar}\right)$ and it satisfies the following relation $T(dx)|x\rangle = |x + dx\rangle$. Then $$\pi T(dx)|x\rangle=\pi |x+dx\rangle=e^{i\delta}|-x-dx\rangle$$ Similarly $$T(-dx)\pi |x\rangle=e^{i\delta}T(-dx)|-x\rangle=e^{i\delta}|-x-dx\rangle$$ This proves eqn (4.2.8). Similarly for eqn (4.2.9) we have: $$\pi T(dx) \pi^{\dagger} |x\rangle=\pi T(dx) (e^{i\delta}|-x\rangle)=\pi (e^{i\delta}|-x+dx\rangle)=|x-dx\rangle=T(-dx)|x\rangle$$ Hence $$\pi \left(1-i\frac{\hat{P}.\hat{dx}}{\hbar}\right)\pi^{\dagger}=\left(1+i\frac{\hat{P}.\hat{dx}}{\hbar}\right)$$

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    $\begingroup$ I dont think dx should be treated as an operator. It should not have a hat. Hence my answer in my comment earlier above - thanks! $\endgroup$ Commented Sep 12, 2021 at 16:41
  • $\begingroup$ This is QM, everything which acts on kets, should be viewed from operator point of view. See, T(dx) is an operator (matrix). P is an hermitian operator. So you can't take dx as an infinitesimal displacement vector, since then P.dx also becomes a vector $\endgroup$
    – paul230_x
    Commented Sep 12, 2021 at 16:48
  • $\begingroup$ I think this is not correct, if $\hat{P}.\hat{dx}$ with both $\hat{P}$ and $\hat{dx}$ are matrices (or operators), do you consider the matrix tensor products between $\hat{P}.\hat{dx}$? (I am not sure we need this here) $\endgroup$ Commented Sep 12, 2021 at 19:42
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    $\begingroup$ I think we only need $\hat{P} \cdot {dx}$ where the $\cdot$ is the $\hat{P}\cdot{dx}=\hat{P^j}\cdot {dx^j}$ where $j$ is the component $j=1,2,3$ of space dimensions, while $dx$ is only a number (thus number of linear algebra) and $\hat{P^j}$ are operators (thus matrix of linear algebra). $\endgroup$ Commented Sep 12, 2021 at 19:44
  • $\begingroup$ So, I was assuming $d\hat{x}$ as some infinitesimal but constant matrices $\{dx_i\}_{i=1,2,..}$. So your 2nd comment is correct. I missed your comment earlier when you said the same thing about treating dx as some infinitesimal constant quantity. Sorry for the unnecessary confusion. Hope that the problem is cleared now. $\endgroup$
    – paul230_x
    Commented Sep 12, 2021 at 20:13

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