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?

  • $\begingroup$ 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, 2017 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, 2017 at 14:40

1 Answer 1


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 $$L^2\left(\mathbb R^3, \frac{d\vec{k}}{2k^0}\right) \ni \psi_{\Phi} \mapsto (2k^0)^{-1/2}\psi_\Phi =: \phi_\Phi \in L^2(\mathbb R^2, d\vec{k}) \:.$$


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