We know that a stationary state is of the form 4[r, t] = 4[r] e¹¹. (1) Here [] is an eigenstate and is the associated eigenenergy. (Remember that ħ= 1 in

natural units.) In the idealized setting normally considered, this is a steady state of the system since it has a time-independent proba- bility density. In practice, though, all excited states have a finite lifetime, 1, and a more realistic representation of the probability density for any excited state is p[t] = e. (2) It is only through QFT that the decay of such "stationary states" are possible. With standard Schrödinger equa- tion quantum mechanics, a pragmatic expedient is to simply adopt a more physically reasonable excited state representation: 4[r, t] = 4[r] e\ อ้ 2 T t [a] Focus on the temporal component of this, T[t] :=et ²1, and calculate the following: (3) (i) Ť:= F[T], the temporal Fourier transform of T[t]. Call the Fourier frequency &, so that you have Ť[ɛ], a complex-valued energy spectrum for the wave function. (ii) The spectral density, D, is defined as D[ɛ] := Ť[ɛ] + Ť[ɛ]* = 2 Re[†[²]. Interestingly, the inverse Fourier transform of D[ɛ] is equal to T[t], so all we have really done is found a real-valued Fourier transform of the time-varying portion of the wave function. If you want, you can test this for yourself by calculating F-¹[D].