4
$\begingroup$

Given an appropriate function space $H$, suppose $H_0$ to be the linear subspace spanned by the solutions of the Klein-Gordon equation and to equip that linear subspace with the inner product

$$\langle \Phi_1 | \Phi_2 \rangle = i\int \mathrm{d}\vec{x}(\Phi_1 ^* \overleftrightarrow{\partial_0}\Phi_2) = i\int \mathrm{d}\vec{x} (\Phi_1 ^* \partial_0\Phi_2 - \Phi_2 \partial_0\Phi_1^*) $$

Question: Are then the multiplicative operators $x^\mu$ hermitian when acting on $(H_0 ,\langle ,\rangle)$? I think that $x^0$ may give some problem.

And if $x^\mu$ are not hermitian, how can one define

$$L_{\mu \nu}=x_\mu i\partial_\nu - x_\nu i \partial_\mu$$

as hermitian generators of a Lorentz group representation over $H_0$?

Finally: with an appropriate choice of $H$, is $(H_0 ,\langle ,\rangle)$ a Hilbert space?

$\endgroup$
  • $\begingroup$ The only $x^\mu$ which makes sense is the Newton-Wigner operator: Newton, T.D.; Wigner, E.P. (1949). "Localized States for Elementary Systems". Reviews of Modern Physics. 21: 400 $\endgroup$ – DanielC Oct 23 '17 at 14:29
  • $\begingroup$ @DanielC Thanks for the comment, but actually I'm not interested in defining a position operator, which is quite a messy problem in QFT. I was just wondering if the multiplicative opertor $x^\mu$ (which I suppose to be well defined even if it may not be so) are hermitian given that linear space and that inner product. I call them "position operator" only because they look like the position operator of non relativistic quantum mechanics $\endgroup$ – L.R. Oct 23 '17 at 14:40
4
$\begingroup$

First of all $H_0$ can be initially taken as the set of the KG solutions of the form $$\Phi(x) = \frac{1}{(2\pi)^{3/2}} \int_{\mathbb R^3} \phi_{\Phi}(\vec{k}) e^{i\left(\vec{k}\cdot \vec{x}-x^0k^0\right)} \frac{d \vec{k}}{\sqrt{2k^0}}\tag{1}$$ with $\phi$ in the space of Schwartz functions and where $$k^0 :=\sqrt{\vec{k}^2 + m^2}\:.$$ With this choice we easily see that $$\langle \Phi_1|\Phi_2\rangle = \int_{\mathbb R^3} \overline{\phi_{\Phi_1}(\vec{k})} \phi_{\Phi_2}(\vec{k}) d\vec{k}$$ Thus the actual Hilbert spece is the completion of $H_0$ with respect to the said scalar product and it is evident that it is isomorphic to $L^2(\mathbb R^3, d\vec{k})$.

With this definition you see that $x^\mu$ is not Hermitian (it is not well defined as its image is outside the Hilbert space: evidently $x^\mu\phi(x)$ is not a KG solution if $\phi$ is in general). However $L_{\mu\nu}$ is Hermtian (and is well defined on the initial said domain). More precisely it is essentially self-adjoint. To prove Hermiticity you just have to pass the operators under the sign of integration integrating by parts. The term $k^0$, which is a function of $\vec{k}$, gives a contribution but all contributions cancel each other in view of the structure of $L_{\mu\nu}$.

ADDENDUM. A unitary equivalent representation is obtained re-defining the Hilbert space using the Lorentz-invariant measure $\frac{d \vec{k}}{2k^0}$ instead of $d \vec{k}$, so that the Hilbert space is $L^2(\mathbb R^3, d\vec{k}/2k^0)$.

With this choice (1) is replaced by $$\Phi(x) = \frac{1}{(2\pi)^{3/2}} \int_{\mathbb R^3} \psi_{\Phi}(\vec{k}) e^{i\left(\vec{k}\cdot \vec{x}-x^0k^0\right)} \frac{d \vec{k}}{2k^0}\tag{2}\:.$$ The unitary map intertwining the two Hilbert space is obviously $$L^2\left(\mathbb R^3, \frac{d\vec{k}}{2k^0}\right) \ni \psi_{\Phi} \mapsto (2k^0)^{-1/2}\psi_\Phi =: \phi_\Phi \in L^2(\mathbb R^2, d\vec{k}) \:.$$

$\endgroup$

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.