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How to calculate phase space shift of function in Dirac Bracket Notation

I'm working through some material in Ch. 2 (p. 70-71) of the book "Ray Tracing and Beyond" by Tracy et. al. on Weyl Symbol Calculus which is the underlying mathematical framework originally developed for quantum mechanics (QM). The authors in the book walk the reader through an example on how to apply the phase space shift operator to a function $\psi(x)$: $$ e^{i(\sigma \hat{x} - \tau \hat{k})} \psi(x), $$ where $\sigma$ and $\tau$ are constants. I would say that they make a range of incorrect arguments, but by coincidence end up at the correct result. My question is whether my method of calculation is correct.

Setting

$\hat{x}, \hat{k}$ are the 1D-position and momentum operators with the canonical commutation relation $[\hat{x}, \hat{k}] = i$ and the momentum eigenstates in the position basis $\left\langle x \middle| k \right\rangle = \frac{1}{\sqrt{2\pi}} e^{i k x}$.

$e^{- i \tau \hat{k}}$ translates in position space, which I prefer to prove by insertion of the identity $\int dk \, \left| k \right\rangle \left\langle k \right|$: $$ \left\langle x \middle| e^{- i \tau \hat{k}} \middle| \psi \right\rangle = \int dk \, \left\langle x \right| e^{- i \tau \hat{k}} \left| k \right\rangle \left\langle k \middle| \psi \right\rangle = \int dk \, e^{- i \tau k} \left\langle x \middle| k \right\rangle \left\langle k \middle| \psi \right\rangle \\ = \int dk \, e^{i k (x - \tau) k} \left\langle k \middle| \psi \right\rangle = \int dk \, \left\langle x - \tau \mid k \right\rangle \left\langle k \middle| \psi \right\rangle = \left\langle x - \tau \middle| \psi \right\rangle $$

Operators act on kets (vectors in the Hilber-space), but Tracy et al. often writes expressions such as $\hat{A} \psi(x)$. I take it to be a fact, that what they really mean is: $\hat{A} \psi(x) \equiv \left\langle x \middle| \hat{A} \middle| \psi \right\rangle$.

Calculation by Tracy et al. on p. 71 in "Ray Tracing and Beyond"

In order to evaulate $e^{i(\sigma \hat{x} - \tau \hat{k})}$, we should use the special case of the Baker-Campbell-Hausdorff Thm. (See Griffiths, p. 121 or Wikipedia): $$ e^{\hat{A} + \hat{B}}=e^{\hat{A}}e^{\hat{B}}e^{- \frac{1}{2}\left[\hat{A}, \hat{B} \right]} $$ We have $\hat{A} = i \sigma \hat{x}$, $\hat{B} = - i \tau \hat{k}$, $\left[\hat{x}, \hat{k} \right] = i$. Therefore I would say that this gives us $\left[\hat{A}, \hat{B} \right] = i \sigma \tau$ and therefore: $$ e^{i(\sigma \hat{x} - \tau \hat{k})} = e^{i \sigma \hat{x}} e^{- i \tau \hat{k}} e^{- \frac{i}{2} \sigma \tau} $$ Mistake 1: But in the book they claim: $$ e^{i(\sigma \hat{x} - \tau \hat{k})} = e^{i \sigma \hat{x}} e^{- i \tau \hat{k}} e^{+ \frac{i}{2} \sigma \tau} $$

Continuing from here, the authors calculates the phase space shift operator by first applying the translation: $$ e^{i(\sigma \hat{x} - \tau \hat{k})} \psi(x) = e^{i \sigma \hat{x}} e^{- i \tau \hat{k}} e^{+ \frac{i}{2} \sigma \tau} \psi(x) = e^{+ \frac{i}{2} \sigma \tau} e^{i \sigma \hat{x}} \psi(x - \tau) $$ Mistake 2: The authors then perform the following specious step, where they let $e^{i \sigma \hat{x}}$ digest the entire argument $x - \tau$: $$ e^{+ \frac{i}{2} \sigma \tau} e^{i \sigma \hat{x}} \psi(x - \tau) = e^{+ \frac{i}{2} \sigma \tau} e^{i \sigma (x - \tau)} \psi(x - \tau) = e^{-\frac{i}{2} \sigma \tau} e^{i \sigma x} \psi(x - \tau). $$ I would argue that this step is also incorrect.

Q3.c with my method and my sign ($e^{- \frac{i}{2} \sigma \tau}$)

If I stick to Dirac bracket notation and the minus sign in the exponent, I actually do end up with the same results as Tracy et al. (but by a very different path!): $$ e^{i(\sigma \hat{x} - \tau \hat{k})} \psi(x) \equiv \left\langle x \right| e^{i \sigma \hat{x}} e^{- i \tau \hat{k}} e^{- \frac{i}{2} \sigma \tau} \left| \psi \right\rangle = e^{- \frac{i}{2} \sigma \tau} e^{i \sigma x} \left\langle x \right| e^{- i \tau \hat{k}} \left| \psi \right\rangle = e^{- \frac{i}{2} \sigma \tau} e^{i \sigma x} \psi(x - \tau) $$

My question: Is it correct that the authors have made two mistakes in their line of arguments? Or is there something I have misunderstood in the Dirac notation used for Weyl Symbol Calculus?

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  • $\begingroup$ Q1, yes; Q2, yours; stop using operators acting on $\psi(x)$ which confuses ordering; simply use the canonical $\hat k\equiv -i\int \!\!dx |x\rangle \partial_x \langle x|$. Q3c is right. $\endgroup$ Commented Feb 18, 2023 at 14:41
  • $\begingroup$ By extension, Q3b is formally right: you should get the wrong answer there. The "digestion" maneuver is specious: $\hat x$ evaluates to plain x in the x-representation, not as a shifted x depending on how you describe a plain function of x ! It appears improvised to cancel one error with another. I recommend you do not call Q1 and Q2 questions, but facts, and focus in assaulting Q3a, in revising your question to be reopened, which is worth doing. Its focus is on daffy cancelling mistakes. $\endgroup$ Commented Feb 18, 2023 at 16:46
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    $\begingroup$ Thank you very much for your swift reply Cosmas. I have revised the question now following your suggestions. $\endgroup$
    – Rune H
    Commented Feb 18, 2023 at 19:03

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Indeed, I agree with your point, and your CBH result/sign (Campbell was first!). The operator $\hat k$ is strictly defined as $$ \hat k = -i\int\!\!dx ~~|x\rangle \partial_x\langle x| ~, $$ which you essentially derived in Fourier space.

Pulling operators outside a matrix element yields the popular expressions for the operator in the x-representation, for the matrix-mechanics averse, $$ \langle x|\hat A | \psi\rangle = \hat A_x \psi(x), $$ where $$ \hat x_x f(x)= x f(x), \\ \hat k_x f(x)= -i\partial_x f(x). $$ The "digest" maneuver of the authors is egregious nonsense, since $\hat k_x$ acts on all x s in the x-representation, and is not "sentient" in assuming special properties in organizing the arguments of the arbitrary function $f(x)$. (This is quite different from the action of $\hat k$ on a bra!!) It appears as though the authors were simply careless and piled up the errors to cancel each other. Caveat lector. Congrats for catching them.

Your Q3.c answer is just fine, and you computed it in the unambiguous recommended manner, in full-fledged Dirac notation, $$ e^{i(\sigma \hat{x} - \tau \hat{k})} \psi(x) \equiv \left\langle x \right| e^{i \sigma \hat{x}} e^{- i \tau \hat{k}} e^{- \frac{i}{2} \sigma \tau} \left| \psi \right\rangle = e^{- \frac{i}{2} \sigma \tau} e^{i \sigma x} \left\langle x \right| e^{- i \tau \hat{k}} \left| \psi \right\rangle = e^{- \frac{i}{2} \sigma \tau} e^{i \sigma x} \psi(x - \tau) . $$ An alternative way to get to the same answer is $$ = \left\langle x \right| e^{-i \tau \hat{k}} e^{i \sigma \hat{x}} e^{\frac{i}{2} \sigma \tau} \left| \psi \right\rangle = e^{ \frac{i}{2} \sigma \tau} \left\langle x -\tau\right| e^{i \sigma \hat{x}} \left| \psi \right\rangle \\ = e^{ \frac{i}{2} \sigma \tau} e^{i \sigma (x-\tau)} \psi(x - \tau) =e^{- \frac{i}{2} \sigma \tau} e^{i \sigma x} \psi(x - \tau) . $$

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    $\begingroup$ Thank you so much Cosmas! It really helps me, that I'm now more sure on how to use the methods correctly. I have accepted your answer. $\endgroup$
    – Rune H
    Commented Feb 19, 2023 at 8:30
  • $\begingroup$ You might enjoy eqn (151) of this. $\endgroup$ Commented Feb 19, 2023 at 20:39

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