Some questions on Invariant Mass Spectrum

[zhangyang]

What can we learn from Invariant Mass Spectrum?How to measure it?So,how to read it?

Mass measurement is converted into energy measurement,but how could we make the quantity change continuously in order to form the horizontal axis?

How to divide different particles to measure them respectively?Is the invariant mass spectrum the unique way to identify a particle?

 

[vanhees71]

The invariant-mass spectrum of what?

One example are particle resonances. In the most simple case the particle decays into two other particles, e.g., a $Z$ boson into an $e^+ e^-$ pair (dilepton). Then you measure the cross section as a function of the invariant mass of the pair. Then you get the $Z$ boson's mass spectrum independent of its motion, because invariant mass is invariant under Lorentz transformations (that's where it's name comes from).

If $p_1$ and $p_2$ are the four-momenta of the two electrons, the invariant mass is given by
$$M^2=(p_1+p_2)^2=p_1^2 + p_2^2 +2 p_1 \dots p_2 = 2 m_e^2 + 2 p_1 \cdot p_2,$$
where the square and the dot product is to be understood as the Minkowski product with the +--- convention (west-coast metric). I also set $c=\hbar=1$ (natural units).

If you plot this spectrum you'll find a peak around the $Z$-boson mass.

 

[zhangyang]

The invariant-mass spectrum of what?

One example are particle resonances. In the most simple case the particle decays into two other particles, e.g., a $Z$ boson into an $e^+ e^-$ pair (dilepton). Then you measure the cross section as a function of the invariant mass of the pair. Then you get the $Z$ boson's mass spectrum independent of its motion, because invariant mass is invariant under Lorentz transformations (that's where it's name comes from).

If $p_1$ and $p_2$ are the four-momenta of the two electrons, the invariant mass is given by
$$M^2=(p_1+p_2)^2=p_1^2 + p_2^2 +2 p_1 \dots p_2 = 2 m_e^2 + 2 p_1 \cdot p_2,$$
where the square and the dot product is to be understood as the Minkowski product with the +--- convention (west-coast metric). I also set $c=\hbar=1$ (natural units).

If you plot this spectrum you'll find a peak around the $Z$-boson mass.

Thank you.

Maybe I haven't considered some aspects,but if it is a two-body-product decay ,in the parent rest frame,the electrons must be back to back,and have same momentum,so how could it generate a spectrum? The inner product would be a constant,right?

 

[vanhees71]

An unstable particle has a finite width in its mass distribution. Otherwise it couldn't decay.

Strictly speaking an "unstable particle" is not a particle in the strict sense, because it cannot be described as an asymptotically free state. So in reality it's the pole of a propagator in a scattering diagram. The spectral distribution is given by the imaginary part of the corresponding retarded propagator (which in vacuum QFT is the same as the usual Feynman propagator for $p^0>0$).

For a scalar particle the propagator takes the general form
$$D(p)=\frac{1}{p^2-m^2-\Pi(p^2)}.$$
If there's a pole in the complex $p^0$ plane and the imaginary part of the self-energy $\Pi$ is not too large, you can write in the neighbourhood of the pole
$$D(p) \simeq \frac{1}{p^2-m^2+m \mathrm{i} \Gamma}.$$
The spectral function then is given by the relativistic Breit-Wigner distribution
$$A(p)=-2 \mathrm{im} D(p)=\frac{2m \mathrm{i} \Gamma}{(p^2-m^2)^2+m^2 \Gamma^2}.$$
This function takes its maximum at $p^2=m^2$ and the distribution falls to half of this value for $p^2-m^2=m \Gamma$. This implies that $\Gamma=1/\tau$ is the inverse lifetime of the particle (measured in inverse MeV, which you can reevaluate to fm with the conversion factor $\hbar c=197 \mathrm{MeV} \; \mathrm{fm}$.

 

[ChrisVer]

Thank you.

Maybe I haven't considered some aspects,but if it is a two-body-product decay ,in the parent rest frame,the electrons must be back to back,and have same momentum,so how could it generate a spectrum? The inner product would be a constant,right?

And do you think that your machine/detector is at the "parent particles rest frame"?
And in general there are many ways this thing cannot help you... For example in LHC where you have parton interactions, you can't possibly know the full momentum of the parton ... also in that case, a Z boson could come from particles that have received radiative correction maybe by the exchange of a gluon (in which case it is even worse to define a rest frame)...
and many other reasons... for example the intermediate particle could not be real... In which case the relation:
$(p_1 + p_2)^2= M_{inv}^2$ is not necessarily taking the value $M_z^2$

 

[mfb]

Most of the time, mass spectra are looked at to distinguish particles (which usually form a peak at their mass, with a finite width due to detector effects and the particle lifetime) from background (sets of particles that do not come from the same decay process and do not show a peak).

 

[zhangyang]

An unstable particle has a finite width in its mass distribution. Otherwise it couldn't decay.

Strictly speaking an "unstable particle" is not a particle in the strict sense, because it cannot be described as an asymptotically free state. So in reality it's the pole of a propagator in a scattering diagram. The spectral distribution is given by the imaginary part of the corresponding retarded propagator (which in vacuum QFT is the same as the usual Feynman propagator for $p^0>0$).

For a scalar particle the propagator takes the general form
$$D(p)=\frac{1}{p^2-m^2-\Pi(p^2)}.$$
If there's a pole in the complex $p^0$ plane and the imaginary part of the self-energy $\Pi$ is not too large, you can write in the neighbourhood of the pole
$$D(p) \simeq \frac{1}{p^2-m^2+m \mathrm{i} \Gamma}.$$
The spectral function then is given by the relativistic Breit-Wigner distribution
$$A(p)=-2 \mathrm{im} D(p)=\frac{2m \mathrm{i} \Gamma}{(p^2-m^2)^2+m^2 \Gamma^2}.$$
This function takes its maximum at $p^2=m^2$ and the distribution falls to half of this value for $p^2-m^2=m \Gamma$. This implies that $\Gamma=1/\tau$ is the inverse lifetime of the particle (measured in inverse MeV, which you can reevaluate to fm with the conversion factor $\hbar c=197 \mathrm{MeV} \; \mathrm{fm}$.

For example, a decay J/psi →p, anti-p,gamma; An article analyze the invariant mass of p bar p as a whole,
"Observation of a Near-Threshold Enhancement in thepp¯Mass Spectrum from RadiativeJ/ψ→γpp¯Decays"(http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.91.022001),
but they treat the whole as a resonance who has a width.What the product of the resonance decay is?And the "whole" is called p bar p ,only because the mass peak is 2m_p?If it is a bound state of p bar p,the static mass would be much smaller because of the bind energy,and particle structure is always not the decay product,because the produce is generated just at the decay time.

How could we change the mass parameter as the horizontal axis in the invariant mass spectrum?

 

[mfb]

I don't understand your questions.

How could we change the mass parameter as the horizontal axis in the invariant mass spectrum?

Which mass parameter do you want to change?
The invariant mass of the p/pbar system is a measurement. For each decay, you calculate that mass, and increase the corresponding bin by 1.

 

[vanhees71]

Again also in this case, you measure the momenta of the outgoing proton and antiproton. Then you know their energies $E_1=\sqrt{m_p^2+\vec{p}_1^2}$ and $E_2 = \sqrt{m_p^2 + \vec{p}_2^2}$. Then you calculate the invariant mass $M^2=(E_1+E_2)^2-(\vec{p}_1+\vec{p}_2)^2$. Since you have an additional particle (a $\gamma$ quantum in that case), you get a broad invariant-mass distribution rather than a sharp peak in this case as you can see in Fig. 3.