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I've been reading through a derivation of the LSZ reduction formula (http://www2.ph.ed.ac.uk/~egardi/MQFT_2013/, lecture 2, pages 2-3) and I'm slightly confused about the arguments made about the assumptions:

$$ \begin{aligned} \langle\Omega\vert\phi(x)\vert\Omega\rangle &=0\\ \langle\mathbf{k}\vert\phi(x)\vert\Omega\rangle &=e^{ik\cdot x} \end{aligned} $$

For both assumptions the author first relates $\phi(x)$ to $\phi(0)$ by using the 4-momentum operator $P^{\mu}$, i.e. $$ \phi(x)=e^{iP\cdot x}\phi(0)e^{-iP\cdot x} $$ such that, in the case of the first assumption, one has $$ \langle\Omega\vert\phi(x)\vert\Omega\rangle =\langle\Omega\vert e^{iP\cdot x}\phi(0)e^{-iP\cdot x}\vert\Omega\rangle =\langle\Omega\vert\phi(0)\vert\Omega\rangle $$ where we have used that the vacuum state satisfies $P^{\mu}\lvert\Omega\rangle =0$, such that $e^{-iP\cdot x}\vert\Omega\rangle = \vert\Omega\rangle$.

What I don't understand is, why do we need to relate $\langle\Omega\vert\phi(x)\vert\Omega\rangle$ to $\langle\Omega\vert\phi(0)\vert\Omega\rangle$ in the first place? Both $\langle\Omega\vert\phi(x)\vert\Omega\rangle$ and $\langle\Omega\vert\phi(0)\vert\Omega\rangle$ are Lorentz invariant.

Is it simply because, by showing that for any $x^{\mu}$, $\langle\Omega\vert\phi(x)\vert\Omega\rangle$ is equal to the Lorentz invariant number, $v\equiv\langle\Omega\vert\phi(0)\vert\Omega\rangle$ (in principle $\langle\Omega\vert\phi(x)\vert\Omega\rangle$ could have a different value for each spacetime point $x^{\mu}$), we can then simply shift the field $\phi(x)\rightarrow \phi(x)-v$, such that the condition $\langle\Omega\vert\phi(x)\vert\Omega\rangle=0$ is satisfied? (If this is the case, then I'm guessing the argument is similar for the second condition.)

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The LSZ formula is based on the following assumptions:

  1. There exists a vector $|\Omega\rangle$ that satisfies $P^\mu|\Omega\rangle=J^{\mu\nu}|\Omega\rangle=0$.

  2. The field transforms according to a certain representation of the Poincaré Group, that is, it satisfies $$ U(a,\Lambda)\phi(x)U(a,\Lambda)^\dagger=D(\Lambda)\phi(\Lambda x+a) $$ where $a\in\mathbb R^4$ and $\Lambda\in SO(1,d)^+$, and $$ U(a,\Lambda)\equiv\mathrm e^{-iP_\mu a^\mu}\mathrm e^{-i\omega_{\mu\nu}J^{\mu\nu}} $$

  3. There exists a certain vector $|\boldsymbol p,\sigma\rangle$ that satisfies $P^\mu|\boldsymbol p,\sigma\rangle=p^\mu|\boldsymbol p,\sigma\rangle$ such that $m^2\equiv p^2$ is an isolated eigenvalue of $P^2$.

  4. The field $\phi(x)$ satisfies $\langle \Omega|\phi(x)|\Omega\rangle=0$.

  5. The field $\phi(x)$ satisfies $\langle \Omega|\phi(x)|\boldsymbol p,\sigma\rangle \neq 0$.

  6. Some other assumptions that are irrelevant for this post (e.g., if the system has a well-defined notion of charge conjugation, then $\phi(x)$ has to commute with $\mathscr C$, and similarly for other internal symmetries).

If $(2)$ is satisfied, and $D(\Lambda)$ is a non-trivial representation of the Lorentz group, then $(4)$ is satisfied automatically; i.e., one need not impose this assumption as a separate condition. Therefore, in this answer we will restrict ourselves to trivial representations of the LG, that is, the scalar representation, where $\phi(x)$ is a scalar field.

In the case of scalar fields, $\langle \Omega|\phi(x)|\Omega\rangle$ is Lorentz invariant regardless of whether it vanishes or not. But we do need to make sure it vanishes, because $(4)$ is a necessary condition for the LSZ formula. Therefore, in order to make sure it vanishes, we note the following: as discussed in the OP, this number satisfies $$ \langle \Omega|\phi(x)|\Omega\rangle=\langle \Omega|\phi(0)|\Omega\rangle $$

Therefore, if for some reason $\langle \Omega|\phi(x)|\Omega\rangle$ is non-zero, we redefine the field $\phi(x)$ through $$ \phi(x)\to\phi(x)-\langle \Omega|\phi(0)|\Omega\rangle $$ which doesn't spoil any of the conditions $1,2,3,5,6$ provided they were already satisfied by the original field, but it ensures that $4$ is satisfied, by construction.

As for the second condition, the argument is as follows: if we use $\langle \Omega|U(a,\Lambda)=\langle \Omega|$ and $U(a,\Lambda)^\dagger|\boldsymbol p\rangle=\mathrm e^{ipa}|\Lambda\boldsymbol p\rangle$, then we can always write $$ \begin{aligned} \langle \Omega|\phi(x)|\boldsymbol p\rangle&=\langle \Omega|\overbrace{U(x,\Lambda)U(x,\Lambda)^\dagger}^1\phi(x)\overbrace{U(x,\Lambda)U(x,\Lambda)^\dagger}^1|\boldsymbol p\rangle\\ &=\overbrace{\langle \Omega|U(x,\Lambda)}^{\langle \Omega|}\overbrace{U^\dagger(x,\Lambda)\phi(x)U(x,\Lambda)}^{\phi(0)}\overbrace{U(x,\Lambda)^\dagger|\boldsymbol p\rangle}^{\mathrm e^{ipx}|\Lambda\boldsymbol p\rangle}\\ &=\langle\Omega|\phi(0)|\Lambda\boldsymbol p\rangle\mathrm e^{ipx} \end{aligned} $$

If we now set $x=0$, we see that this implies that $$ \langle\Omega|\phi(0)|\boldsymbol p\rangle=\langle\Omega|\phi(0)|\Lambda\boldsymbol p\rangle $$ i.e., the matrix element $\langle\Omega|\phi(0)|\boldsymbol p\rangle$ is a scalar; but the only scalar function of $\boldsymbol p$ is $p^2=m^2$, and therefore this matrix element is just a constant, independent of $\boldsymbol p$: $$ \langle \Omega|\phi(x)|\boldsymbol p\rangle=c\, \mathrm e^{ipx} $$

Finally, if, as in $(5)$, we assume that $\langle \Omega|\phi(x)|\boldsymbol p\rangle \neq 0$, then $c\neq 0$ and we can always redefine $\phi(x)$ so that $c=1$; and, as again, this doesn't spoil any of the conditions $1,2,3,4,6$.

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  • $\begingroup$ Thanks for the detailed answer. So is the point that one can show that $\langle\Omega\vert\phi(x)\vert\Omega\rangle$, whose value could in principle be different at each spacetime point $x^{\mu}$, is actual equal to the constant $\langle\Omega\vert\phi(0)\vert\Omega\rangle$ $\forall\;x^{\mu}$, and as such we can simply shift the field at each spacetime point to $\phi(x)-\langle\Omega\vert\phi(0)\vert\Omega\rangle$?! $\endgroup$ – user35305 Feb 13 '17 at 14:20
  • $\begingroup$ @user35305 yes, that is correct. Note also that there is nothing special about $x=0$; we could also shift the field to $\phi(x)-\langle\Omega|\phi(\infty)|\Omega\rangle$ or any other point, because $\langle\Omega|\phi(x)|\Omega\rangle$ is independent of $x$. Therefore, we can set $x$ to any other convenient value. $\endgroup$ – AccidentalFourierTransform Feb 13 '17 at 14:29
  • $\begingroup$ I'm guessing this is the reason why one translates $\langle\Omega\vert\phi(x)\vert\mathbf{p}\rangle$ and $\langle\Omega\vert\phi(0)\vert\mathbf{p}\rangle$ as well?! Like you put in your answer, by carrying out this procedure one can show that $\langle\Omega\vert\phi(x)\vert\mathbf{p}\rangle$ is a Lorentz scalar and so we can again use this information to redefine $\phi(x)$ to ensure that $\langle\Omega\vert\phi(x)\vert\mathbf{p}\rangle =e^{ip\cdot x}$. $\endgroup$ – user35305 Feb 13 '17 at 14:40

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