0

I am reading the "Quantum Field Theory lectures of Sidney Coleman". In the first chapter (subchapter 1.2), the author talks about translation invariance. In particular, he states that $U(a)=e^{iP\cdot a}$ is the translation operator, with $$O(x+a)=U(a)O(x)U^{\dagger}(a)$$ where $O$ is any other operator depending on $x^{\mu}$. Then, the author reduces the translations to space translations and therefore $$e^{-i\textbf{P}\cdot\textbf{a}}O(\textbf{x})e^{i\textbf{P}\cdot\textbf{a}}= O(\textbf{x}+\textbf{a}).$$ But then, the author states that only operators localized in space transform according to this rule, giving a counter-example the operator $\hat{\textbf{q}}$ $$e^{i\textbf{P}\cdot\textbf{a}}\hat{\textbf{q}}e^{-i\textbf{P}\cdot\textbf{a}}= \hat{\textbf{q}}+\textbf{a}.$$

I have three questions:

  1. What is the underlying reason for only operators localized in space being transformed under this rule?

  2. Why is $\hat{\textbf{q}}$ an operator that is not localized in space?

  3. How do the operators that are not localized in space transform?

Any help will be appreciated.

Qmechanic
  • 201,751
schris38
  • 3,950
  • Related: https://physics.stackexchange.com/q/747455/247642 – Roger V. Jan 30 '23 at 08:08
  • So if $\hat{\textbf{q}}$ is an operator that is not localized in space because it can take any real value, what is there from stopping me in assuming that every $O(\textbf{x})$ is such an operator @RogerVadim? – schris38 Jan 30 '23 at 08:22
  • 1
    There's also the correspondence principle - your operators should give classical physics as a limiting case. But p and q are not necessarily momentum and position. E g., this could electric and magnetic field, charge and superconducting phase, etc. – Roger V. Jan 30 '23 at 08:54

1 Answers1

2

  1. Consider the following example of an operator "localized in space": The orthogonal projection operator $$\Pi_\delta(x) = \int\limits_{x-\delta}^{x+\delta} \! \! \! dq \, |q \rangle \langle q |, \quad {\rm where} \quad Q | q \rangle = q | q \rangle, $$ corresponds to the observable of finding the particle in the interval $[x-\delta, x+\delta]$ centered around $x$. I have assumed only a single spatial dimension for simplicity (the generalization to three spatial dimensions is trivial). $Q$ denotes the position operator. Using $e^{-iPa} |q \rangle = |q+a \rangle$ (c.f. second line of eq. (1.39) in the book), one obtains $$e^{-iPa} \Pi_\delta(x) e^{iPa}= \! \int\limits_{x-\delta}^{x+\delta} \! \! \! dq \, |q+a \rangle \langle q+a| =\! \! \!\! \int\limits_{x+a-\delta}^{x+a+\delta} \! \! \! \! dq \, | q\rangle \langle q | = \Pi_\delta(x+a)$$ in accordance with the third line of eq. (1.39).

  2. $Q$ itself does not single out a specific point in space. You could take e.g. the somewhat pathological "operator" $\delta(Q-x)= |x\rangle \langle x|$, corresponding to the limiting case of $\Pi_\delta(x)/2 \delta$ for $\delta \to 0$.

  3. The canonical commutation relation $[ Q, P] = i \mathbf{1}$ implies $e^{-iPa} f(Q) e^{iPa} = f(Q-a)$ and $e^{-iPa} g(P) e^{iPa}= g(P)$.

Hyperon
  • 5,748
  • I do not see how (1) and (2) answer my corresponding questions. Also, I fail to see why $\delta(Q-x)=|x\rangle\langle x|$ (where is the $Q$ dependence on the RHS??). And as far as (3) is concerned, clearly the book states that $\hat{\textbf{q}}$ transforms as $e^{i\textbf{P}\cdot\textbf{a}}\hat{\textbf{q}}e^{-i\textbf{P}\cdot\textbf{a}}=\hat{\textbf{q}}+\textbf{a}$, which is not covered by neither $f(Q)$, nor $g(P)$, if I am not mistaken... – schris38 Jan 30 '23 at 12:22
  • @schris38 1. If you want to construct an "operator localized in space" (in the sense of the book), you have to construct a function of the operators $Q$ (and/or $P$) that singles out a specific point $x$ in space (or, more generally, a region in space (i.e. an operator of the general form $F(Q, P, x)$). The operators $\Pi(x)$ and $\delta(Q-x)$ given above above are just two simple examples (of infinitely many possibilities). 2. Using the definition of a function of an operator (spectral theorem), you find $\delta(Q-x)= \int dq \delta(q-x) |q \rangle \langle q |= |x \rangle \langle x |$. – Hyperon Jan 30 '23 at 13:23
  • @schris38 3. $e^{-iPa} Q e^{iPa} = Q-a$ is a special case of $e^{-iPa} f(Q) e^{iPa} =f(Q-a)$ for the function $f(x) =x$. – Hyperon Jan 30 '23 at 13:27
  • What does it mean "to single out a specific point in space"? Isn't your $Q$ a special case of $F(Q,P,x)$, and hence an operator that should transform as $O(x)$? Also, a question on the notation: what is the relation between the eigenvalue of $Q$ on the position basis and $x$? – schris38 Jan 30 '23 at 13:38
  • i think you need a normalizing factor $\frac{1}{2\delta}$ in front of the integral to make it a "pathological" operator. – hyportnex Jan 30 '23 at 15:04
  • @schris38 I should have been more explicit on this point. Forgetting the possible depence on the momentum operator $P$, the meaning is to choose an $x$-dependent operator of the form $F(Q,x) =f(Q-x)$. Both operators in the examples fulfill this requirement: $\Pi(x) = c_I(Q-x)$, where $c_I$ is the characteristic function of the interval $[-\delta,\delta]$, and obviously also $\delta(Q-x)$. The spectrum of $Q$ is $\mathbb{R}$ and also $x \in \mathbb{R}$. – Hyperon Jan 30 '23 at 15:11
  • @hyportnex Thank you, I have corrected it! – Hyperon Jan 30 '23 at 15:31