# How to understand complex masses of unstable particles? The conceptual problem of calculating decay rate

If a particle has a complex mass, $$p^2-m^2=0$$ leads to $$p^μ \notin \mathbb R^4$$. What does it mean?

When you want to calculate S-matrix elements of decay process $$\langle p_f,\ldots\mid p_i\rangle$$, you compute the $$n$$-point correlation function $$\langle 0\mid Tφ(x_1)φ(x_2)\ldots\mid 0\rangle$$ and using the LSZ formula. Fourier transformation of the correlation function has the poles at $$p_j^2=m_j^2$$ ,because it includes the vacuum polarization amplitudes$$= \frac i {p_j^2-m_j^2+O((p_j^2-m_j^2)^2)}\approx \frac i{p_i^2-m_i^2}$$ on each external legs in diagrams. The LSZ formula says when you multiply $$Π_j(p_j^2-m_j^2)/i$$ and take the limit $$p_j^2→m_j^2$$, you get S-matrix elements $$\langle p_f,\ldots\mid p_i\rangle$$. Therefore, you may know that you need to consider only "amputated diagrams" when you calculate S-matrix.

However, beyond the tree level, the vacuum polarization amplitude of unstable particle has the pole at not real but complex value. This means on shell limit $$p_i^2→m_i^2$$ leads to $$p_i^\mu \notin \mathbb R^4$$. On the other hand, if you take limit such as $$p_i^2\to\operatorname{Re} m_i^2$$, you cannot do the approximation "vacuum polarization amplitudes $$\approx \frac i{p_i^2-m_i^2}$$", because $$p_i^2$$ is far from the pole.

After all, my question is how to compute decay rates of unstable particles correctly. Can I calculate S-matrix elements by computing "amputated diagrams" as usually done?

The related problem is noted in Srednicki's textbook on P.162 http://web.physics.ucsb.edu/~mark/qft.html
He said in and out state should consist of infinitely long-lived particles. He thought unstable particles present as intermediate states, and regarded decay late as a quantity related to the width of resonance. (p.165 (25.25))

$$f(E)=\frac{1}{E-E_0+iΓ/2}.$$

Fourier transformation of $$f(E)$$ gives $$g(t)=\exp(iE_0t-Γt/2)$$, which seems something expressing decay. I want to know too the precise meaning of this procedure. (At first, I considered the time evolution of $$|ψ\rangle = \int dE \, f(E)|E\rangle$$ by Schrodinger equation, but $$|ψ(t)\rangle\neq g(t)|ψ(0)\rangle$$).

I'll discuss this in the context of Yukawa theory, and use renormalized perturbation theory. The way I understand it is this.

Setup: Consider the renormalization scheme

$$\psi_R = \frac{1}{\sqrt{Z}}\psi_0 \\ m_R = \frac{1}{Z_m} m_0$$

where $$Z_i = 1 + \delta_i$$. Then the renormalized propagator is.

$$G^R = \frac{1}{Z_2} G^{bare}.$$

Now recall that summing over all 1PI insertions we may write

$$iG^{bare} = \frac{i}{\not{p} - m_0 + \Sigma(\not{p})}$$

where $$\Sigma$$ is the sum over all 1PI insertions. Now, we can write $$\Sigma = \Sigma_2 + {\cal O}(g^4)$$ keeping only those insertions of order less than $$g^4$$. So that our renormalized propagator to order $$g^2$$ is

$$G^R = \frac{1}{1+\delta}\frac{i}{\not{p} - m_0 + \Sigma_2(\not{p})} \\ \boxed{G^R= \frac{i}{\not{p} - m_R + \Sigma_R(\not{p})} }$$

where we have defined

$$\Sigma_R = \Sigma_2 + \delta \not{p} - (\delta + \delta_m)m_R$$

Now, recall that we define the physical mass as the pole of the propagator. Therefore,

$$m_P - m_R + \Sigma_R(m_P) = 0 .$$

This implies that

$$\boxed{\delta_m m_P = \Sigma_R(m_P)}$$

Your question: Let us call $$\Sigma_2(\not{p}) = i\Delta m$$, anticipating that the correction is purely imaginary. Then, our propagator is

$$\tilde{G}^R= \frac{i}{\not{p} - m_R+ i\Delta m}$$

where now we know that our propagator has a pole at $$m_P$$.

Recall that the amplitude for a particle to propagate from $$x$$ to $$y$$ is given by the fourier transform

$$D(x-y) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{\not{p} - m_R + i\Delta m}e^{-ip(x-y)}$$

To make things easier suppose that y=0. Then we have

$$D(x) = \int \frac{d^4p}{(2\pi)^4} \frac{i}{\not{p} - m_R+ i\Delta m}e^{i\mathbf{p\cdot x}}e^{-iEt}$$

Where $$E = \sqrt{p^2 + m_{rest}^2}$$. What is the mass of the particle in its rest frame? Well by definition it's the pole of the propagator! And so we may replace

$$E = \sqrt{p^2 + (m_R - i \Delta m)^2} \\ =\sqrt{\left(-{\Delta m}^2+m^2+p^2\right)^2+4 {\Delta m}^2 m^2} \left(i \sin \left(\frac{\phi}{2}\right)+\cos \left(\frac{\phi}{2}\right)\right)\\ \boxed{E \equiv \xi \left(i \sin \left(\frac{\phi}{2}\right)+\cos \left(\frac{\phi}{2}\right)\right) }$$

where $$\phi$$ is the $$\mathrm{Arg}$$ of the radicand.Finally, we obtain that

$$\boxed{D(x) = \int \frac{d^4p}{(2\pi)^4} \frac{ie^{i\mathbf{p\cdot x}} e^{-i\xi\cos(\frac{\phi}{2}) t}}{\not{p} - m_R+ i\Delta m}e^{-\xi\sin(\frac{\phi}{2}) t}}$$

and so we see that indeed the probability for a particle to exist at a time $$t$$ decays exponentially.

• Thank you for answering my questions! I clearly understand how to interpret decay rates. D(y-x)=<0|Tφ(y)φ(x)|0>=<$y_i|exp(-iH(y_0-x_0)|x_i$>, therefore the probability amplitude of the state will exponentially decrease as time goes on according to your calculation. I did not suppose Fourier transformation of f(E) appeared in the calculating D(y-x) from $\tilde G^R(p)$. I greatly appreciate your answer! – Takumi Hayashi Jan 9 '19 at 7:29
• Right! That’s how I think about it at least – InertialObserver Jan 9 '19 at 7:44