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I have been going over this derivation of the Breit-Wigner formula for resonance in particle physics but cannot reconcile the steps with my knowledge of QM.

The initial state is given by:

$$ \psi(t)=\psi(t=0)e^{-iE_0t}e^{-\frac{t}{2\tau}}$$

Here arises my first question:

  1. Is dependence on position neglected? If so, why?

Then, it is stated

$$\textrm{Prob}(\textrm{ find state } |\psi\rangle)\propto e^{-\frac{t}{\tau}} $$

  1. Finding the state $|\psi\rangle$ where? At time $t$? What does this mean?

We can now convert this to the energy domain by Fourier transforming this $\psi(t)$:

$$f(E)=\int_0^\infty \textrm{d}t\,\psi(t)e^{iEt}$$

and we get

$$f(E)= \dfrac{i\psi(0)}{(E_0-E)-\frac{i}{2\tau}}$$

  1. Why is this a Fourier transform if the range starts at $0$ and not at $-\infty$?
  2. Why is this valid? I am used to converting from position to momentum space, but time-energy is something I have never done in QM.
  3. Moreover, what are the time eigenstates? For position and momentum we have $|x\rangle$ and $|p\rangle$, but for time?

The procedure then goes on and asserts that the probability of finding the state $|\psi\rangle$ with energy $E$ is given by

$$|f(E)|^2=\dfrac{|\psi(0)|^2}{(E_0-E)^2+\frac{1}{4\tau^2}} $$

  1. Shouldn't it be $|f(E)|^2\textrm{d}E$?
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    $\begingroup$ On your question 5, time has no eigenstates in the QM established originally. As such, this is not a good question if you ignore contributions to foundational QM arrived at since ~1983. $\endgroup$
    – PrawwarP
    Commented Dec 5, 2020 at 13:58
  • $\begingroup$ 1,2 Everywhere, so anywhere in space. Integrate over all positions. $\endgroup$ Commented Dec 5, 2020 at 14:57
  • $\begingroup$ 3 Because infinite negative t would get ψ to explode; formally, you could hack-extend it for a full transform of the Cauchy distribution, of course, with an absolute value. $\endgroup$ Commented Dec 5, 2020 at 23:52
  • $\begingroup$ @CosmasZachos with respect to 1,2, what do you mean by integrate over all positions? Integrate exactly what? I also do not get how your answer to 3 justifies it being a Fourier transform, and also do not know how it explains what this step is doing $\endgroup$ Commented Dec 6, 2020 at 13:59
  • $\begingroup$ @PrawwarP with all due respect, your answer is not helping me in anyway, but thank you anyway for letting me know there are no time eigenstates $\endgroup$ Commented Dec 6, 2020 at 14:02

1 Answer 1

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I fear one is shadow-boxing with your undisclosed text. All good QM texts cover this, but one doesn't know what you are taking issue with. The state is $$ \psi(t)=\psi(0)~e^{-iE_0t}e^{-\frac{t}{2\tau}},$$ so the probability of it not having decayed is monotonically decreasing, $$ |\psi(t)|^2 / |\psi(0)|^2 = e^{-t/\tau}, $$ the standard exponential decay law. Could multiply with the number of such particles to get a bulk survival probability, e.g. of a chunk of radioactive material.

(1,2) Any conceivable space dependence has been integrated out, since it is irrelevant to the decay. The state could be anywhere and everywhere in space, and its decay would not be affected by space considerations--think of doing all the space integrals in advance. The square of the wave function, then, is a probability of existence, in the whole universe, of that state, and not a probability space-density. Note the state is a hamiltonian eigenstate, but the eigenvalue is not real, $E_0-i/2\tau$, because the hamiltonian is not hermitian. The probability of the existence of the state as a fraction of an initial probability of 1, when you start measuring time, is thus decreasing all the way to 0 at infinite time.

(3) Your time range is then [0,$\infty$), and that is what you integrate over, so you are only doing half a Fourier transform, since the full Fourier transform would take you back to an infinite value (duh!), and you only wish to monitor survival probability relative to a starting time 0.

(4) Valid? it is a formal operation: $$f(E)=\int_0^\infty \textrm{d}t\,\psi(t)e^{iEt} = \dfrac{i\psi(0)}{(E-E_0)+\frac{i}{2\tau}} ~,$$ giving you a spectral decomposition of your state, and is useful in the undisclosed applications of your text. It is essentially the propagator of the unstable state in question, providing the amplitude for the decay.

(6) Indeed, normally $|f(E)|^2$ would correspond to a probability density in E, a Lorentzian, or Cauchy distribution, whose (full) FT, as you see, gives you an $\propto e^{-|t|/\tau}$, half of which you have been using here.

(5) is obscure... Time is a parameter.

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  • $\begingroup$ Thank you so much for your help. Regarding (1,2), as you say, position has been integrated out. If we have a wavefunction $\psi(x,t)$ and we only care aabout its $t-$dependence, can we just integrate over position? What is the physical significance of doing this? $\endgroup$ Commented Dec 8, 2020 at 19:45
  • $\begingroup$ Also, sorry if I did not phrase the question properly, but by valid in (4) I meant if this process has some physical interpretation. It's more or less the same question as in my first comment $\endgroup$ Commented Dec 8, 2020 at 19:47
  • $\begingroup$ Yes, the x dependence of ψ is some distribution of finite extent centered someplace. If we integrate it in x, we get an amp that "normally" should square-integrate to 1: the particle is assuredly someplace in the universe, and we don't care where. (But the decay will decrease the probability to less than one, and, eventually, 0.) $\endgroup$ Commented Dec 8, 2020 at 20:13
  • $\begingroup$ I'm not sure about physical interpretation; our physical intuition is informed by monkey-see-monkey-do math. It is some sort of non relativistic limit of a propagator in a persistence amplitude as a function of energy... The imaginary part is the "width" Γ eating up probability of survival... $\endgroup$ Commented Dec 8, 2020 at 20:17
  • $\begingroup$ Okay, thank you very much for your answer. I'll try to reflect upon the physical intuition a bit more $\endgroup$ Commented Dec 9, 2020 at 15:21

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