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].