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In Sidney Coleman's Lectures he talked about space translations such that

$$\tag{1} e^{ia P}\rho(x) e^{-ia P} ~=~ \rho(x-a),$$

but when I expanded the exponentials and used the commutation relation of $P=-i\frac{d}{dx}$ and $x$, I got

$$\tag{2} e^{ia P}\rho(x) e^{-ia P} ~=~ e^{ia [P,~\cdot~]}\rho(x) ~=~e^{a\frac{d}{dx}}\rho(x)~=~\rho(x+a)$$

with a plus instead of $\rho(x-a)$ with a minus as Coleman (1).

This was on Coleman's lectures on QFT (1975-76), Lecture 3, about 12:20 minutes into the lecture.

Does anyone see where I may have gone wrong?

References:

  1. Coleman's lectures on QFT (1975-76), Lecture 3, about 12:20 minutes into the lecture.

  2. S. Coleman, Notes from Sidney Coleman's Physics 253a, arXiv:1110.5013, p. 23.

Qmechanic
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MrDi
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2 Answers2

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This is indeed confusing, as the signs change depending on whether you act on states or on operators.

Thus if you have a state $|\Psi\rangle$ and operate on it with the translation $ U=e^{-ipa}$ you get $$ \langle x|e^{-ipa}|\Psi\rangle = e^{-a\frac d{dx}}\langle x|\Psi\rangle =\sum_{n=0}^\infty\frac{(-a)^n}{n!}\frac{d^n\langle x|\Psi\rangle}{dx^n}=\langle x-a|\Psi\rangle. $$ (Here I've implemented the relation $``p=-i\frac d{dx}\text'\text'$ as its more rigorous form $\langle x|p=-i\frac d{dx}\langle x|$, which translates the above into the form $\langle x|e^{-ipa}=e^{-a\frac d{dx}}\langle x|=\langle x-a|$.)

If you're considering expectation values, then, you want to calculate something like $\langle \Psi'|f(\hat x)|\Psi'\rangle$, where $f$ is any function (with the hat added here only, for clarity) and $|\Psi'\rangle=U|\Psi\rangle$. Such an expectation value can be written as $$ \langle\Psi|U^\dagger f(x)U|\Psi\rangle =\int\text dx\langle \Psi|U|x\rangle f(x)\langle x|U|\Psi\rangle=\int\text dx\langle \Psi|x-a\rangle f(x)\langle x-a|\Psi\rangle =\langle\Psi|f(x+a)|\Psi\rangle. $$ This is also what you get from the operator algebra, which you have done correctly.

There is one circumstance in which this transforms differently, though. If $\rho(x)$ is the probability density, then it really means the diagonal elements of the density matrix, $$\rho(x)=\rho(x,x)=\langle x|\Psi\rangle\langle\Psi|x\rangle.$$ In this case, the term $e^{ipa}\rho(x)e^{-ipa}$ is really a shorthand way of making $U$ act on the state, seen in the position basis, so that $$ e^{ipa}\rho(x)e^{-ipa} :=\langle x|U^\dagger|\Psi\rangle\langle\Psi|U|x\rangle =\langle x-a|\Psi\rangle\langle\Psi|x-a\rangle =\rho(x-a) $$ as in Coleman's claim.

Added:

What Coleman is actually doing is similar to the above, but not quite. For him $\rho(x)$ is the electron density operator at position $x$. This means that $x$ is here just a number and $\rho(x)$ is an operator; when he says it's "just a delta function" he means an expression of the form $$\rho(x)=\delta(\hat x-x),$$ or a sum of such terms if there is more than one electron to contend with. In my view, this delta function expression overcomplicates things, and this can be expressed much more simply as $$\rho(x)=|x\rangle\langle x|$$ (or a sum of such terms), which you can get from the above by inserting $1=\int\text dx' |x'\rangle \langle x'|$.

It is clear now that in this instance $\rho(x)$ must transform as a state, and not an operator. You can see this as $$ \langle x'|e^{ipa}|x\rangle =\langle x'+a|x\rangle =\langle x'|x-a\rangle \quad\text{so}\quad e^{ipa}|x\rangle=|x-a\rangle, $$ and similarly on the conjugate, or you can see it as $\langle x|e^{-ipa}=\langle x-a|$ left-multiplied by its conjugate.

The relationship to the calculation above is interesting: Coleman is interested in the physical quantity $\langle \Psi|\rho(x)|\Psi\rangle$, and this is equal to the diagonal elements of the density matrix, $\langle x|\Psi\rangle\langle \Psi|x\rangle$, which is treated above. Of course, both quantities must transform equally under the same translation.

Emilio Pisanty
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I would like to add some precisions to Emilio Pisanty's answer, and show that you have to be cautious with the notation $O(x)$, when $O$ is an operator.

The translation $x \to x'=x +a$, induces a transformation in the space of states, which is is : $$|\psi\rangle \to |\psi'\rangle = e^{-ia.P}|\psi\rangle \tag{1}$$

This corresponds to the transformation :

$$\langle x |\psi\rangle \to \langle x |\psi'\rangle = \langle x - a|\psi\rangle \tag{2}$$ That is :

$$\psi(x) \to \psi'(x) = \psi(x-a) \tag{2a}$$

An operator $O$ transforms as :

$$O \to O' = e^{-ia.P} ~ O~e^{ia.P} \tag{3}$$ You can check that all this is coherent and that the object $O|\psi \rangle$ transforms as a state.

The transformation $(3)$ corresponds to the transformation :

$$\langle x |O| y\rangle \to \langle x |O'| y\rangle \tag{4} = \langle x -a |O| y -a\rangle$$ That is :

$$O(x,y) \to O'(x,y) = O(x-a,y-a)\tag{4a}$$

Now, if you compare your first formula with the formula $(3)$, you see a sign inversion, so, starting with your formula (which corresponds to a translation $x \to x - a$), we will have :

$$O(x,y) \to O'(x,y) = O(x+a,y+a)\tag{4b}$$ which is true for any operator, so it is true for the density operator. Of course, you may be interested only by the diagonal terms, with $x=y$, but it is only a special case of the formula $4b$, that is :

$$O(x,x) \to O'(x,x) = O(x+a,x+a)\tag{4b'}$$

So, if you understand all this, you may write, in a short notation $O(x) = O(x,x)$, but if you have some difficulties, it is better not to use it, and to use notations $O,\langle x |O| y\rangle, O(x,y), O(x,x)$

So, your calculus is correct.

Trimok
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