Klein–Gordon Inner Product

The Klein–Gordon inner product is a conserved pairing of solutions, not the ordinary spatial integral of two amplitudes. On the full complex solution space it is indefinite. On a chosen free positive-frequency subspace it becomes positive and defines a one-particle Hilbert space. Keeping the normalization and the subspace choice explicit prevents confusion between negative norm, negative frequency, and negative probability.

Required background. The Conserved Current supplies the scalar bilinear; Plane-Wave Solutions fixes its Fourier coefficients; Relativistic Currents provides hypersurface integration and boundary conditions.

Polarizing the conserved current

Let $m>0$ and let $\phi_1,\phi_2$ solve the same Klein–Gordon equation with the same charge and real prescribed background. Polarization replaces the repeated amplitude in the current by two different solutions:

$$j^\mu[\phi_1,\phi_2]=\frac{i\hbar}{2m} \left[\phi_1^*D^\mu\phi_2-(D^\mu\phi_1)^*\phi_2\right].$$

The conservation proof applies without change to this possibly complex current. For a spacelike Cauchy surface with future orientation, define

$$(\phi_1,\phi_2)_{\rm KG} =\frac1c\int_\Sigma j^\mu[\phi_1,\phi_2]\,d\Sigma_\mu.$$

Assume sufficient decay or boundary conditions that eliminate the flux through the intervening spatial boundary. The divergence theorem then makes the result independent of $\Sigma$. It is antilinear in its first argument, linear in its second, and Hermitian: $(\phi_1,\phi_2)_{\rm KG}=(\phi_2,\phi_1)_{\rm KG}^*$.

On a constant-time slice the formula becomes

$$\begin{aligned} (\phi_1,\phi_2)_{\rm KG} ={}&\frac{i\hbar}{2mc^2}\int d^3x \left(\phi_1^*\partial_t\phi_2 -(\partial_t\phi_1)^*\phi_2\right)\\ &-\frac{q}{mc^2}\int d^3x\,\Phi\phi_1^*\phi_2. \end{aligned}$$

This pairing is gauge invariant because both solutions transform with the same phase. In the rest of the mode calculation take $A_\mu=0$.

Orthogonal frequency sectors

Use the Fourier coefficients $a,b$ of the plane-wave owner, whose two branches both carry spatial phase $e^{i\mathbf p\cdot\mathbf x/\hbar}$. The spatial integral produces $(2\pi\hbar)^3\delta^3(\mathbf p-\mathbf q)$. For two positive-frequency factors the time derivative supplies $(E_{\mathbf p}+E_{\mathbf q})/(2mc^2)$; two negative-frequency factors supply its negative. The mixed derivative factor is proportional to $E_{\mathbf p}-E_{\mathbf q}$ and vanishes on the delta function. Hence

$$(\phi_1,\phi_2)_{\rm KG} =\int d^3p\,\frac{E_{\mathbf p}}{mc^2} \left(a_1^*a_2-b_1^*b_2\right).$$

This proves integrated positivity on the positive-frequency subspace; it does not assert pointwise positivity of the density. The local counterexample is derived in Probability-Density Problems.

Define delta-unit modes by

$$u_{\mathbf p}^{\pm}(x) =\frac{1}{(2\pi\hbar)^{3/2}} \sqrt{\frac{mc^2}{E_{\mathbf p}}} e^{i\mathbf p\cdot\mathbf x/\hbar} e^{\mp iE_{\mathbf p}t/\hbar}.$$

Then

$$\begin{aligned} (u_{\mathbf p}^+,u_{\mathbf q}^+)_{\rm KG}&=\delta^3(\mathbf p-\mathbf q),\\ (u_{\mathbf p}^-,u_{\mathbf q}^-)_{\rm KG}&=-\delta^3(\mathbf p-\mathbf q),\\ (u_{\mathbf p}^+,u_{\mathbf q}^-)_{\rm KG}&=0. \end{aligned}$$

If $\phi=\int d^3p\,(A u^+ +B u^-)$, coefficient extraction is $A(\mathbf p)=(u_{\mathbf p}^+,\phi)_{\rm KG}$ and $B(\mathbf p)=-(u_{\mathbf p}^-,\phi)_{\rm KG}$. The minus sign is essential: the negative-frequency basis is not an orthonormal basis of a positive metric space.

Null vectors do not make the pairing degenerate

A solution with equal positive and negative sector norms can have $(\phi,\phi)_{\rm KG}=0$ without being zero. Every real free solution is an example. This means the form is indefinite, not that the solution is orthogonal to every other solution.

Indeed, on initial data $f=\phi(0)$ and $h=\partial_t\phi(0)$ the free pairing is proportional to $\int(f_1^*h_2-h_1^*f_2)d^3x$. If this vanishes for all independently variable test data $f_2,h_2$, then both $f_1$ and $h_1$ vanish, so the solution is zero. Subject to the stated solution and boundary spaces, the form is nondegenerate. Quotienting out all zero-norm vectors would therefore be invalid; those vectors do not even form a linear subspace in general.

A positive Hilbert space requires a choice

Set $b=0$ and complete suitable positive-frequency solutions in

$$\|\phi\|_{\rm KG}^2=\int d^3p\, \frac{E_{\mathbf p}}{mc^2}|a(\mathbf p)|^2.$$

Multiplying $a$ by $\sqrt{E_{\mathbf p}/(mc^2)}$ identifies this space unitarily with ordinary $L^2(d^3p)$. Proper orthochronous Lorentz transformations preserve the positive massive shell; with scalar field transformation and the conserved pairing they act unitarily on this space. The raw Fourier coefficient is not itself a Lorentz scalar.

In time-dependent backgrounds the full pairing remains conserved, but a preferred positive-frequency splitting need not exist. If suitable asymptotic stationary regions define “in” and “out” splittings, comparing them is extra dynamical information. A conserved indefinite form does not guarantee that a chosen positive subspace is preserved by that comparison or that a particle-creation probability has already been computed.

Translating common normalizations

In natural units, many field-theory texts use $i\int d^3x\,\phi_1^*\overleftrightarrow{\partial_t}\phi_2$. This is $2m$ times the form used here. Their delta-unit modes consequently have factor $1/\sqrt{2E_{\mathbf p}}$, rather than $\sqrt{m/E_{\mathbf p}}$. Changing both the form and the mode factors preserves the stated orthogonality relations. Combining the form from one convention with modes from the other does not. For a massless field, use a normalization without division by $m$; the massive normalization here has no direct $m=0$ substitution.

Exercises

1. In a periodic box of volume $V$, normalize a positive-frequency mode to KG norm one using the current convention of this page.

Solution

The mode is $\sqrt{mc^2/(EV)}e^{-iEt/\hbar+i\mathbf p\cdot\mathbf x/\hbar}$. Its density is $1/V$, so its integral is one. Box momentum values are discrete and mode orthogonality uses a Kronecker delta.

2. Let $(u,u)=1$, $(v,v)=-1$, and $(u,v)=0$. Show that $u+v$ is null but is not orthogonal to $u-v$.

Solution

$(u+v,u+v)=1-1=0$, whereas $(u+v,u-v)=1+1=2$. The zero self-pairing does not erase the vector from the solution space.

3. In a static electrostatic potential $V=q\Phi$, let two normalizable stationary modes have distinct real energies $E_1,E_2$ and vanishing boundary flux. Derive their weighted orthogonality condition.

Solution

Their pairing is

$$e^{i(E_1-E_2)t/\hbar}\int d^3x\, \frac{E_1+E_2-2V(\mathbf x)}{2mc^2} u_1^*(\mathbf x)u_2(\mathbf x).$$

Conservation and $E_1\ne E_2$ force the integral to vanish. Ordinary unweighted $L^2$ orthogonality does not follow from this argument.

References

• N. D. Birrell and P. C. W. Davies, Quantum Fields in Curved Space, Cambridge University Press, 1982, doi:10.1017/CBO9780511622632 — conserved scalar pairings and frequency choices.
• H. Feshbach and F. Villars, “Elementary Relativistic Wave Mechanics of Spin 0 and Spin 1/2 Particles,” Reviews of Modern Physics 30, 24–45, 1958, doi:10.1103/RevModPhys.30.24 — the indefinite scalar charge form.
• D. Tong, Quantum Field Theory, University of Cambridge lecture notes, 2006–2007, §2, Free Fields — field-theory mode and state normalizations.