1
$\begingroup$

In Peskin & Schroeder's book, the authors say that if $\phi(x)$ is the (quantum) Klein-Gordon field operator in Schrödinger's picture and $|0\rangle$ is the vacuum state, the application $$\phi(x)|0\rangle = \int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{2E_{p}}e^{-ip\cdot x} |p\rangle $$ is interpreted as creation of a particle at position $x$. The argument is that, apart from the factor $1/2E_{p}$ in the above formula, the expression is exactly the eigenstate of the position $|x\rangle$ for non-relativistic QM.

Can someone help me understand this interpretation?

My points:

(a) Aren't the eigenstates of the position operator $x$ delta distributions? The integral on the right hand side of the above expression resembles the wave function of the free particle, not the eigenstates of the position to me.

(b) The authors say that $\langle x | p\rangle = e^{ip\cdot x}$ helps the aforementioned interpretation. How come?

$\endgroup$
1
1
$\begingroup$

In non-relativistic QM, generalized eigenstates of the position operator are delta distributions in the position basis, but are plane waves in the momentum basis, i.e. $|x\rangle = \int \mathrm dp \ e^{-ipx} |p\rangle$. See e.g. here, trying to ignore the rather unpleasant typsetting in the second half of the page.

As for $(b)$, note that precisely the same relation holds in nonrelativistic QM, see e.g. here.

$\endgroup$
3
  • $\begingroup$ Why is the factor of 1/2E_p there or any web source you know of that explains it (I couldn't understand the Peskin Schroeder approach) $\endgroup$ – Shashaank Feb 13 at 7:25
  • 1
    $\begingroup$ @Shashaank it's required to make the integration measure Lorentz-invariant: see physics.stackexchange.com/q/83260 and links therein $\endgroup$ – Nihar Karve Feb 13 at 9:02
  • $\begingroup$ It's there becuase the plane wave $|p\rangle$ is normalized so that $\langle p|p'\rangle= (2\pi)^3 2 E_p\delta^3(p-p')$. This is natural as the particle density increases with $E$ because of Lorentz contraction. $\endgroup$ – mike stone Feb 13 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.