A QFT Srednicki Chap 6: Weyl Ordering

TeethWhitener

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In Srednicki’s QFT Chapter 6 (intro to path integrals), he introduces Weyl ordering of the quantum Hamiltonian:
$$H(P,Q)=\int{\frac{dx}{2\pi}\frac{dk}{2\pi} e^{ixP+ikQ}}\int{dp \text{ }dq\text{ }e^{-ixp-ikq}H(p,q)}$$
where ##P,Q## are momentum and position operators and ##H(p,q)## is the classical Hamiltonian. He then mentions that the matrix element for a single time slice of the path integral ##\langle q_2 | e^{-iH\delta t}|q_1\rangle## is given by:
$$\langle q_2 | e^{-iH\delta t}|q_1\rangle =\int{\frac{dp_1}{2\pi}e^{-iH(p_1,\overline{q}_1)\delta t}e^{ip_1(q_2-q_1)}}$$
where ##\overline{q}_1=\frac{1}{2}(q_1+q_2)##. I’m having trouble showing how the midpoint ##\overline{q}_1## enters into the equation. I expanded the propagator to first order and inserted a complete set of momentum states to get:
$$\langle q_2 | e^{-iH\delta t}|q_1\rangle =\int{dp_1\langle q_2|p_1\rangle\langle p_1|q_1\rangle}-\int{dp_1}\int{\frac{dx}{2\pi}\frac{dk}{2\pi}\langle q_2|e^{ixP}|p_1\rangle\langle p_1|e^{ikQ}|q_1\rangle} \times \int{dp\text{ }dq \text{ }e^{-ixp-ikq}H(p,q)}$$
Letting ##\langle q|p\rangle =(2\pi)^{-1/2}\exp{(ipq)}##, this simplifies to:
$$\langle q_2 | e^{-iH\delta t}|q_1\rangle =\int{\frac{dp_1}{2\pi}e^{ip_1(q_2-q_1)}}\left(1-i\delta t\int{\frac{dx}{2\pi}\frac{dk}{2\pi}e^{ixp_1+ikq_1}}\int{dp\text{ }dq\text{ }e^{-ixp-ikq}H(p,q)}\right)$$
It looks like if you just let the ##\exp{(ixp_1)}## integral become two delta functions ##\delta(p_1)## and ##\delta(x_1)##, you’ll get a matrix element dependent on the Hamiltonian ##H(p_1,q_1)## instead of the Hamiltonian at the position midpoint ##H(p_1,\overline{q}_1)##. What am I missing?
 

samalkhaiat

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What am I missing?
The operator [itex]\pi (\alpha , \beta ) \equiv e^{i( \alpha P + \beta Q )}[/itex] factors as [tex]\pi ( \alpha , \beta ) = e^{- \frac{i}{2}\alpha \beta} \ e^{i \alpha P} \ e^{i \beta Q } .[/tex] You can also show that [tex]\pi ( \alpha , \beta ) = \int dx \ e^{i \beta x } \ |x - \frac{\alpha}{2} \rangle \langle x + \frac{\alpha}{2}|.[/tex] So, [tex]\pi (\alpha , \beta ) |q_{2}\rangle = e^{ i \beta (q_{2} - \frac{\alpha}{2}) } \ | q_{2} - \alpha \rangle .[/tex] Thus [tex]\langle q_{1}| \pi ( \alpha , \beta ) | q_{2} \rangle = e^{i \beta (q_{2} - \frac{\alpha}{2}) } \ \delta (q_{1} - q_{2} + \alpha ) = e^{i \beta \ ( \frac{q_{1} + q_{2}}{2} ) } \ \delta (q_{1} - q_{2} + \alpha) . \ \ \ \ \ (1)[/tex] Now, take the matrix element [itex]\langle q_{1}| \hat{H}| q_{2}\rangle[/itex] of the Weyl operator [tex]\hat{H} = \frac{1}{ (2 \pi )^{2}} \int_{\mathbb{R}^{4}} d \alpha \ d \beta \ dq \ dp \ e^{- i( \alpha p + \beta q )} \ H(q,p) \ \pi ( \alpha , \beta ) ,[/tex] and use (1). Doing the [itex]\beta[/itex]-integral first gives you [itex]2 \pi \delta \left(q - \frac{q_{1} + q_{2}}{2} \right)[/itex]. Then, the q-integral gives you [itex]H\left(\frac{q_{1} + q_{2}}{2} , p \right)[/itex]. And finally the [itex]\alpha[/itex]-integral gives you [tex]\langle q_{1}| \hat{H}| q_{2}\rangle = \frac{1}{2 \pi} \int dp \ e^{i (q_{1} – q_{2})p} \ H \left( \frac{q_{1} + q_{2}}{2} , p \right) .[/tex]
 

TeethWhitener

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Thank you so much for your always insightful answers. Baker-Campbell-Hausdorff always seems to trip me up. I’ll work through this tomorrow with a paper and pencil and let you know if I have more questions.
 

samalkhaiat

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Thank you so much for your always insightful answers.
Thank you, it’s my pleasure. The matrix element of Weyl operator (the last equation in #2) brings back old (under graduate) memories. I derived it during my struggle to prove the classical limit [itex]\lim_{\hbar \to 0} \frac{1}{i\hbar} [ \hat{H} , \hat{A}] = \{ H , A \}_{PB}[/itex]. I recall my professor telling me that the proof require advanced mathematics and that I should wait until I study group theory and learn about the star product. But, I asked myself, if the classical limit of Schrodinger equation does not require “fancy mathematics”, then why should Heisenberg equation be any different? Anyway, with a bit of hard work, I managed to do it using my under graduate knowledge in math and QM. When I showed my professor (the late David Bohm) the proof, he went through it then looked at me and said “Son, you are crazy. Ask the secretary to photocopy it for me.”
 

Demystifier

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When I showed my professor (the late David Bohm)
David Bohm was your professor? Wow! May I ask what is your opinion on Bohmian mechanics?
 

samalkhaiat

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David Bohm was your professor? Wow!
Yes, he taught QM1 at Birkbeck College even though he was Emeritus Professor at that time. Bohm was great as a man, brilliant as a teacher and I cannot find the appropriate words to describe his sharp intuitive mind. I used to love the way he obtained the values of very complicated integrals without doing any integration. Unfortunately, he stopped teaching when I started my post-graduate study.

May I ask what is your opinion on Bohmian mechanics?
Well, he never anticipated the name. I learnt about Bohm’s 1952 paper from his friend Basil Hiley who taught us “Advanced Topics in QM”. My first reaction was, and still is, the fact that QM loses its universal property of being representation-independent theory. I recall asking Basil Hiley “If I write Schrodinger equation in (say) the momentum representation, can you derive for me the two Bohm’s equations?” His answer was “No, but many of my PhD students are working on it.”
 

Demystifier

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My first reaction was, and still is, the fact that QM loses its universal property of being representation-independent theory.
I am not trying to turn it into a yet another Bohmian mechanics thread, but let me just note that Newtonian mechanics also "loses" the universal representation independence property that Hamiltonian mechanics has, and yet nobody complains about that. Positions and momenta are physically different, even if some convenient formalisms treat them similarly.
 

Demystifier

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@samalkhaiat What is special about Weyl ordering in comparison with other orderings? In what sense is this ordering better than other orderings?

And one additional question. Would it be correct to say that, effectively, path integral quantization first orders operators at non-equal times according to time ordering and then orders operators at the same time according to Weyl ordering? (By effectively, I mean path integral quantization does not deal with operators explicitly, but is equivalent to operator quantization with the given ordering.)
 

samalkhaiat

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let me just note that Newtonian mechanics also "loses" the universal representation independence property that
True, and it is exactly this ugly feature of Newton mechanics that led Hamilton and others to their extremely useful and symmetric formulation of mechanics on symplectic manifold.
 

samalkhaiat

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@samalkhaiat What is special about Weyl ordering in comparison with other orderings? In what sense is this ordering better than other orderings?
Symmetry is beautiful and powerful concept. This is why the symmetric ordering of Weyl is far more elegant and useful than other ordering. It leads naturally to correct expression of the path integral.

And one additional question. Would it be correct to say that, effectively, path integral quantization first orders operators at non-equal times according to time ordering and then orders operators at the same time according to Weyl ordering? (By effectively, I mean path integral quantization does not deal with operators explicitly, but is equivalent to operator quantization with the given ordering.)
If I understood you correctly, then yes. In the path integral, you do the Weyl ordering at each time slice.
 

Demystifier

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Symmetry is beautiful and powerful concept. This is why the symmetric ordering of Weyl is far more elegant and useful than other ordering.
Maybe it's elementary, but in what sense is the Weyl ordering symmetric? Do you mean symplectic symmetry or some other symmetry?
 

samalkhaiat

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Maybe it's elementary, but in what sense is the Weyl ordering symmetric? Do you mean symplectic symmetry or some other symmetry?
It is symmetric in both sense of the word! Weyl ordering is symmetric in the literal sense, i.e., ##\hat{\pi} (xp) = \frac{1}{2} \left( \hat{x}\hat{p} + \hat{p}\hat{x} \right)##, as well as in the sense of preserving the symplectic structure.
 
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TeethWhitener

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Terence Tao has a relatively easy-to-follow blog post on Weyl quantization:
Terence Tao said:
...there is one standard choice for a functional calculus available for general operators
$$A_1,\ldots,A_k$$
, namely the Weyl functional calculus; it is analogous in some ways to normal coordinates for Riemannian manifolds, or exponential coordinates of the first kind for Lie groups, in that it treats lower order terms in a reasonably nice fashion. (But it is important to keep in mind that, like its analogues in Riemannian geometry or Lie theory, there will be some instances in which the Weyl calculus is not the optimal calculus to use for the application at hand.)
 

Demystifier

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Weyl ordering is symmetric in the literal sense, i.e., ##\hat{\pi} (xp) = \frac{1}{2} \left( \hat{x}\hat{p} + \hat{p}\hat{x} \right)##,
Can it be used as an alternative definition of Weyl ordering (as the ordering that is symmetric under permutations) that replaces the usual definition in terms of a Fourier transform?
 

samalkhaiat

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Can it be used as an alternative definition of Weyl ordering (as the ordering that is symmetric under permutations) that replaces the usual definition in terms of a Fourier transform?
That would be an insult to Weyl. There are far more to Weyl’s quantization than just symmetrizing product of operators. It led to: the theory of Pseudo differential operators also known as Weyl calculus; functional analysis on locally compact groups; deformation quantization with the star product; ##C^{\ast}## algebras; non-commutative geometry and the so-called Wigner phase-space QM.
 

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