# single-particle wavepackets in QFT and position measurement

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Consider a scalar field $\phi$ described by the Klein-Gordon Lagrangian density $L = \frac{1}{2}\partial_\mu \phi^\ast\partial^\mu \phi - \frac{1}{2} m^2 \phi^\ast\phi$.

As written in every graduate QM textbook, the corresponding conserved 4-current $j^\mu = \phi^\ast i \overset{\leftrightarrow}{\partial^\mu} \phi$ gives non-positive-definite $\rho=j^0$. If we are to interpret $\phi$ as a wave function of a relativistic particle, this is a big problem because we would want to interpret $\rho$ as a probability density to find the particle.

The standard argument to save KG equation is that KG equation describes both particle and its antiparticle: $j^\mu$ is actually the charge current rather than the particle current, and negative value of $\rho$ just expresses the presence of antiparticle.

However, it seems that this negative probability density problem appears in QFT as well. After quantization, we get a (free) quantum field theory describing charged spin 0 particles. We normalize one particle states $\left|k\right>=a_k^\dagger\left|0\right>$ relativistically:

$$\langle k\left|p\right>=(2\pi)^3 2E_k \delta^3(\vec{p}-\vec{k}), E_k=\sqrt{m^2+\vec{k}^2}$$

Antiparticle states $\left|\bar{k}\right>=b_k^\dagger \left|0\right>$ are similarly normalized.

Consider a localized wave packet of one particle $\left| \psi \right>=\int{\frac{d^3 k}{(2\pi)^3 2E_k} f(k) \left| k \right>}$, which is assumed to be normalized. The associated wave function is given by

$$\psi(x) = \langle 0|\phi(x)\left|\psi\right> = \int{\frac{d^3 k}{(2\pi)^3 2E_k} f(k) e^{-ik\cdot x}}$$

$$1 = \langle\psi\left|\psi\right> = \int{\frac{d^3 k}{(2\pi)^3 2E_k} |f(k)|^2 } = \int{d^3x \psi^\ast (x) i \overset{\leftrightarrow}{\partial^0} \psi (x)}$$.

I want to get the probability distribution over space. The two possible choices are:

1) $\rho(x) = |\psi(x)|^2$ : this does not have desired Lorentz-covariant properties and is not compatible with the normalization condition above either.

2) $\rho(x) = \psi^\ast (x) i \overset{\leftrightarrow}{\partial^0} \psi(x)$ : In non-relativistic limit, This reduces to 1) apart from the normalization factor. However, in general, this might be negative at some point x, even if we have only a particle from the outset, excluding antiparticles.

How should I interpret this result? Is it related to the fact that we cannot localize a particle with the length scale smaller than Compton wavelength ~ $1/m$ ? (Even so, I believe that, to reduce QFT into QM in some suitable limit, there should be something that reduces to the probability distribution over space when we average it over the length $1/m$ ... )

This post imported from StackExchange Physics at 2015-05-16 18:38 (UTC), posted by SE-user pdfs
asked Nov 3, 2013
retagged May 16, 2015
In your mind, what does mean the notation $\psi^\ast i \overset{\leftrightarrow}{\partial^0} \psi$ ? If you think that , here, $\psi$ is a state $|\psi\rangle$, this is a nonsense, because $|\psi\rangle$ does not depends on $t$. If you think that $\psi$ is an operator, this is a nonsense too, because t the integration on $x$ gives an operator too, and it cannot be equals to $1$, which is a real number, but not an operator.
Yes, I do. By substituting $\psi(x) = \int{\frac{d^3 k}{(2\pi)^3 2\omega_k} f(k) exp(-ik\cdot x)}$ , you can directly verify the equation you pointed out.. By the way, how can I use LaTeX in comment lines?