# Relation between decay rate (lifetime) and spectral width

1. Jun 19, 2009

### Chen

Hi,

I am trying to understand exactly how and why the lifetime, or decay rate, of an atomic level determines the spectral width of the transition to this level. Also I would like to understand why the natural lineshape is a Lorentzian. I am familiar with the vague explanations involving Fermi's golden rule and time-energy uncertainty, but I am looking for a more convincing description.

Thanks,
Chen

2. Jun 19, 2009

### malawi_glenn

why is Fermis Golden rule vauge? Do you want a Quantum Field Theory calculation or what?

3. Jun 19, 2009

### Chen

Well, doesn't it give the wrong result for the lineshape (sinc and not Lorentzian)?

4. Jun 19, 2009

### malawi_glenn

ok I give you a quick derivation then:

the time-dependent excited state wavefunction, with lifetime tau and energy E_f:

$$\Psi _f (\vec{r},t) = \Psi _f (\vec{r},0)\exp (-iE_f(t/\hbar) - t/\tau)$$

Fourier transform:

$$\Psi _f (\vec{r},0)\exp (-iE_f(t/\hbar) - t/\tau) = \int \frac{\hbar / \tau }{ (E_f-E)^2 + (\hbar/\tau)^2}e^{-iE_f(t/\hbar)}dE$$

thus the energy spectrum has a Lorentzian shape

5. Jun 19, 2009

### Chen

Thanks, although that's not derived from Fermi's golden rule is it...

I was hoping for a more intuitive explanation. Pressure broadening stems from enhanced stimulated emission, doppler broadening is obvious... where does the natural linewidth comes from?

6. Jun 19, 2009

### malawi_glenn

Not it is not, just basic QM.

What is intuitive and not is relative, this is going from time to Energy, why is not a good old Fourier transform intuitive enough? It was you who wanted the real explanation, well here it is... it is just basic QM, it comes from the formalism itself -> Energy (the Hamiltonian) is the generator of Time evolution.

The intuition must be within the realm where the physics is taking place, this is quantum physics, one must therefore develop intuition for Fourier transforms etc.

7. Jun 19, 2009

### Chen

Ok. Maybe someone else can provide a different explanation.

8. Jun 19, 2009

### malawi_glenn

what do you want? The real deal or a fairy tale? Why is intuition always linked to classical physics? I mean, I think it is good that you did not accept the Energy-Time uncertainty relation, because that is such fairytale. The most convincing explanation must be the correct one, don't you think?

(I am very interested in how physics students think of and reason around QM, that is why I ask these questions)

9. Jun 19, 2009

### Chen

If I knew what I wanted to hear I wouldn't need to ask you for it.

Can you please give others a chance to answer my question as well? I thank you for your answer but I want to hear other possible answers.

10. Jun 19, 2009

### malawi_glenn

You asked "exactly how and why the lifetime, or decay rate, of an atomic level determines the spectral width of the transition to this level" and "Also I would like to understand why the natural lineshape is a Lorentzian." You called Fermi golden rule and E-t uncertainty relation VAUGE, and you are saying that the explanation I gave you was not "intuitive", clearly I should ask what is wrong?

11. Jun 19, 2009

### Bob S

Ask an electrical engineer about the Fourier transform from the time domain to the frequency domain. If you have a limited time dt sample of a line frequency w (=2 pi f), the measured linewidth is dw. In particular, dt dw ~=1. Now multiply both sides by h-bar (Planck's constant) and get dE dt ~= h-bar.. So the uncertainty principle is not unique to quantum mechanics.

12. Jun 19, 2009

### Avodyne

The case of an atomic level is quite complicated, because it requires quantum field theory for the emitted photon. Much easier is potential scattering with a resonance. This is discussed in most textbooks (eg, Merzbacher). The atomic-level case is essentially the same; we can think of the decay as a scattering process that starts with the atom in an unstable state, and ends with the atom in a stable state (or perhaps a different unstable state) plus an emitted photon. Then there is a pole in the S-matrix element in the complex energy plane near the real axis; this is what fixes both the Lorentzian line shape and the relation between width and lifetime. Again, this is explained for potential scattering in most QM textbooks.

13. Jun 20, 2009

### granpa

I know nothing about quantum mechanics but even a passing knowledge of Fourier transforms makes it obvious that the shorter the transition the wider the line width.

14. Jun 20, 2009

### malawi_glenn

Avodyne answered the question I gave you in my first post "Do you want a Quantum Field Theory calculation or what? " But you did not answered it ;-)

Now when you have got more replies, you have to ask yourself what is the most intuitive one:
i) Spectral width vs. time due to the energy-time relation which makes it possible to perform Fourier transforms.
ii) Poles in the complex energy plane in the transition S-matrix =D

15. Jun 20, 2009

I do not know if this is of any help, but I would simply think of a two-level system. Let the energies of the levels are $$\epsilon_1$$ and $$\epsilon_1$$. Without any hybridization, the spectral function for the first level is simply
$$A(1,\omega) = -\frac{1}{\pi} \textrm{Im} G_{11}^R \propto \delta (\omega-\epsilon_1)$$.
For hybridized states one gets
$$G_{11}^R (\omega) = (\omega + i \eta - \epsilon_1 - \frac{|t|^2}{\omega + i \eta - \epsilon_2})^{-1} \approx (\omega + i \eta - \epsilon_1 + i \Gamma/2)^{-1}$$.
This leads to the spectral function of Lorentzian shape:
$$A(1,\omega) \propto \frac{\Gamma}{(\omega-\epsilon_1)^2+(\Gamma/2)^2}$$.
Usually the real part of the self-energy is absorbed into the energy and the imaginary part is responsible for broadening (in this case the broadening may not be weakly energy-dependent, of course). Broadening is also reflected in the fact that the pole of the Green's function is at $$z=\epsilon_1 - i\Gamma/2$$, away from the real axis. The imaginary part then gives a finite life-time (damping) in time domain, i.e. $$\tau = (\Gamma/2)^{-1}$$.

16. Jun 20, 2009

### conway

1. I have to ask if you understand why the lineshape is lorentzian for a classical LC circuit or an ordinary antenna? If so, then your question might be rephrased: why should we expect the physics of the atom to be similar to these classical cases? This is something we could discuss further.

2. I would also like someone to present a working definition of what we mean by "natural" lineshape. There is obviously a "practical" lineshape you could measure by doing spectroscopy on a sample of hot glowing gasses, but this surely not what we are talking about here?