# Unitary spacetime translation operator

Srednicki writes: We can make this a little fancier by defining the unitary spacetime translation operator

$$T(a) \equiv \exp(-iP^\mu a_\mu/ \hbar)$$

Then we have $$T(a)^{-1} \phi(x) T(a) = \phi(x-a)$$

How do we get the second equation from the first equation?

-
I suspect Mr Taylor (of series fame) might be able to help with this..? – twistor59 Jun 29 '13 at 19:02

It is unnecessary to use Hilbert space states to formally derive this result; it quickly follows from a useful result about the matrix exponential (which comes in very handy when one studies Lie algebras which, incidentally, is essentially what we're looking at here).

Let $X$ be any $n$-by-$n$ complex matrix, then we define the linear operator $\mathrm{ad}_X$ on the vector space of such matrices by $$\mathrm{ad}_X Y = [X,Y]$$ for all $n$-by-$n$ complex matrices $Y$. Here $[\cdot, \cdot]$ denotes the commutator often called the adjoint operator. Then we have the following result: $$e^XYe^{-X} = e^{\mathrm{ad}_X}Y$$ Now, if we formally apply this result to linear operators on the Hilbert space of a quantum field theory, then we obtain $$T(a)^{-1}\phi(x)T(a) = e^{ia_\mu P^\mu/\hbar}\phi(x)e^{-ia_\mu P^\mu/\hbar} = e^{\mathrm{ad}_{ia_\mu P^\mu/\hbar}}\phi(x) = \sum_{k=0}^\infty \frac{\mathrm{ad}^k_{ia_\mu P^\mu/\hbar}\phi(x)}{k!}$$ Now we use the fact that for any field $\Phi$, we have $$\mathrm{ad}_{ia_\mu P^\mu/\hbar}\phi(x)=\frac{ia_\mu}{\hbar} [P^\mu, \phi(x)] = \frac{ia_\mu}{\hbar}(i\hbar \partial^\mu)\phi(x) = -a_\mu \partial^\mu\phi(x)$$ Applying this result $k$ times and inserting it into the series expansion for the exponential written above, we obtain $$T(a)^{-1}\phi(x)T(a) = \sum_{k=0}^\infty\frac{(-1)^k}{k!}(a_\mu\partial^\mu)^k\phi(x)$$ Now, we simply note that the right hand side is the Taylor expansion of $\phi(x-a)$. Explictly $$\phi(x-a) = \phi(x) -a_\mu\partial^\mu\phi +\frac{1}{2}(a_\mu\partial^\mu)^2\phi(x) + \cdots + \frac{(-1)^k}{k!}(a_\mu\partial^\mu)^k\phi(x)+\cdots$$ and this gives the desired result.

-
I am reading QFT from Srednicki's book. In the 2nd chapter of this book and in the spin half part of this book, group theory and group representation theory is used. Can you suggest me a book from where I can learn this? – omephy Jun 30 '13 at 12:44
I personally like Brian Hall's Lie Groups, Lie Algebras, and Representations – joshphysics Jun 30 '13 at 18:52
Thnaks a lot. I have benn waiting for your answer. Srednicki is killing me! – omephy Jun 30 '13 at 18:53
This might be helpful too for references physics.stackexchange.com/questions/6108/… – joshphysics Jun 30 '13 at 18:58
I also like Group Theory in Physics by Wu-Ki Tung. – joshphysics Jun 30 '13 at 18:58

[EDIT] Supposing a state basis $|e_i\rangle$, we are going to use the following notation for a state: $|A(x)\rangle = \sum_i A_i(x)|e_i\rangle$

An operator $O$ appying on $A$ gives then: $O|A(x)\rangle = \sum_i OA_i(x)|e_i\rangle$

For instance, $\partial_{\mu}|A(x)\rangle = \sum\partial_{\mu}A_i(x)|e_i\rangle$

Now, instead of working with operators, I think it is simpler to work with states $\left|A(x)\right\rangle$ and $\left|B(x)\right\rangle$ such as:

$$\left|B(x)\right\rangle = \Phi(x) \left|A(x)\right\rangle \tag{1}$$

This is true, of course, for $x-a$, that is:

$$\left|B(x - a)\right\rangle = \Phi(x- a) \left|A(x-a)\right\rangle \tag{2}$$

We know that:

$$P_{\mu}\left|A(x)\right\rangle = i \hbar \partial_{\mu}\left|A(x)\right\rangle.$$

So, we get: \begin{align} T(a)^{-1}\left|A(x)\right\rangle &= e^{\large iP_\mu a^\mu/ \hbar}\left|A(x)\right\rangle\\ &=e^{\large -a^{\mu}\partial_{\mu}} \left|A(x)\right\rangle\\ &= \left|A(x - a)\right\rangle \end{align}

The last equality is simply the Taylor series of $\left|A(x - a)\right\rangle$ at $x$, that is:

$$\left|A(x - a)\right\rangle = \left|A(x)\right\rangle - a^{\mu}\partial_{\mu} \left|A(x)\right\rangle + \frac{1}{2!} (a^{\mu}\partial_{\mu})^2\left|A(x)\right\rangle +\frac{(-1)^n}{n!}(a^{\mu}\partial_{\mu})^n\left|A(x)\right\rangle + \dots.$$

Now, applying $T(a)^{-1}$ to equation $(1)$, we get:

$$T(a)^{-1}\left|B(x)\right\rangle =T(a)^{-1}\Phi (x)\left|A(x)\right\rangle.$$

That is:

$$T(a)^{-1}\left|B(x)\right\rangle =T(a)^{-1}\Phi (x)T(a)T(a)^{-1}\left|A(x)\right\rangle.$$

So, we get:

$$\left|B(x - a)\right\rangle =T(a)^{-1}\Phi (x)T(a)\left|A(x - a)\right\rangle.$$

Looking at equation $(2)$, we finally get:

$$T(a)^{-1}\Phi (x)T(a) = \Phi (x-a)$$

-
nooooooooo! twistor59 was trying to lead OP to this own solution :( – joshphysics Jun 29 '13 at 19:58
@Trimok: Thanks a lot. You have written: $e^{-a.\partial} |A(x) \rangle$. Here what does $a.\partial$ mean? – omephy Jun 29 '13 at 20:00
@Ome: $a.\partial$ means $a^{\mu}\partial_{\mu}$. I made an edit in the answer. – Trimok Jun 29 '13 at 21:03
@Ome : Note that, with your convention $T(a) \equiv \exp(-iP_\mu a^\mu/ \hbar)$, I then use the convention $P_{\mu}\left|A(x)\right\rangle = i \hbar \partial_{\mu}\left|A(x)\right\rangle.$. But other conventions are $T(a) \equiv \exp(iP_\mu a^\mu/ \hbar)$ and $P_{\mu}\left|A(x)\right\rangle = -i \hbar \partial_{\mu}\left|A(x)\right\rangle.$. This does not change the final result. – Trimok Jun 29 '13 at 21:24
@Trimok What precisely is the definition of the state $|A(x)\rangle$ that you're using? In particular, what is the mathematical context of your first equation? – joshphysics Jun 30 '13 at 2:37