I QFT for the Gifted Amateur Question?

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1. Sep 25, 2016

43arcsec

In chapter 9 "Quantum Mechanical Transformations", example 9.3, can anyone explain how $$\hat{p}$$ in the exponential converts itself to q? Apologies in advance if this is very basic, but thanks for looking.

$$\hat{u}(a)|q>=e^{-i\hat{p}\cdot a} |q>$$

$$\hat{u}(a)|q>=e^{-iq\cdot a} |q>$$

2. Sep 25, 2016

Fightfish

I assume $|q\rangle$ is a momentum eigenstate? Then its just a matter of using $\hat{p}|q\rangle = q |q\rangle$

3. Sep 25, 2016

dextercioby

If that would be the case, then it's highly unfortunate to use q for labeling momentum in nonrelativistic QM. QM stays somewhere at the middle between Hamiltonian CM and QFT. In Hamiltonian CM "q" stands for generalized coordinate, while in QFT q goes along with p,r,s in Perturbation theory and Feynman diagrams to label 4-momenta. In "ordinary" QM, the momentum operator is p with a hat or without, its eigenket is |p> and its eigenvalue in the space of eigendistributions is also p. Why use q at all ?

4. Sep 25, 2016

43arcsec

Wow, thanks for the quick replies. Fightfish is correct: |q> is a momentum state. That was stated in the example; sorry, I should have included it. Unfortunately, I am still a bit confused. Why is it OK to write $$\hat{p} \quad for \quad e^{-i\hat{p}\cdot a}$$ in the footnote 4 of chapter 8, Lancaster writes in general:
$$e^{\hat{A}}=1+\hat{A}+\frac{1}{2!}\hat{A}\hat{A}+...$$
That seems a bit messier.

5. Sep 25, 2016

Fightfish

That is the definition of the exponential of an operator.
It's not - $\hat{p}$ is not the same as $e^{-i\hat{p}\cdot a}$. When we act $e^{-i\hat{p}\cdot a}$ on the momentum eigenstate, what we really are doing is to use that expansion, which is in powers of $\hat{p}$, act it on the momentum eigenstate, and then collect the terms back together to obtain another exponential, this time in terms of the eigenvalue.

6. Sep 26, 2016

mbond

One has:
$$e^{-i\hat{\textbf{p}}\textbf{a}}|\textbf{q}\rangle=[1+(-i\hat{\textbf{p}}\textbf{a})+\frac{1}{2!}(-i\hat{\textbf{p}}\textbf{a})^2+\cdots]|\textbf{q}\rangle$$
$\hat{\textbf{p}}=-i\vec{\nabla}$ so it commutes with $\textbf{a}$ then, with $\hat{\textbf{p}}|\textbf{q}\rangle=\textbf{q}|\textbf{q}\rangle$, one has:
$$(-i\hat{\textbf{p}}\textbf{a})|\textbf{q}\rangle=-i\textbf{qa}|\textbf{q}\rangle$$
$$(-i\hat{\textbf{p}}\textbf{a})^2|\textbf{q}\rangle=(-i\textbf{qa})^2|\textbf{q}\rangle$$
$$e^{-i\hat{\textbf{p}}\textbf{a}}|\textbf{q}\rangle=[1+(-i\textbf{qa})+\frac{1}{2!}(-i\textbf{qa})^2+\cdots]|\textbf{q}\rangle=e^{-i\textbf{qa}}|\textbf{q}\rangle$$

7. Sep 26, 2016

43arcsec

Fightfish and mbond, thank you so much for your explanations; they perfectly filled in the steps I was missing.