# 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.