# 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.

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user Trimok
@Trimok By \psi(x), I meant the associated wave function <0|phi(x)|psi>, which is a c-number-valued field defined in the spacetime. (I think it is quite a common practice to use the same letter when we write the abstract state ket and its wave function in QM ...) The scalar field operator is written as \phi(x) in my notation.

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user pdfs
So, you mean $\psi(x)$ in the notation $\psi^\ast i \overset{\leftrightarrow}{\partial^0\psi}$ ?

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user Trimok
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?

This post imported from StackExchange Physics at 2015-05-16 18:39 (UTC), posted by SE-user pdfs