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In section 134 of Vol. 3 (Quantum Mechanics), Landau and Lifshitz make the energy complex in order to describe a particle that can decay:

$E = E_0 - \frac{1}{2}i \Gamma$

The propagator $U(t) = \exp(-i H t)$ then makes the wavefunction die exponentially with time. But also, $H$ is non-Hermitian.

My question: Do we have to modify the basic postulates of quantum mechanics (as described by Shankar, say, or the earlier sections of Landau & Lifshitz) to describe unstable particles?

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Hi user22037 - I took out the explicit "reference request" part of your question. Answerers can still point you to books or papers etc. if they think they'll be useful, but it's usually better if you don't ask for them explicitly so that someone can also just post an explanation. (Also, this sort of thing isn't what the reference-request tag is for.) –  David Z Apr 5 '13 at 23:34

3 Answers 3

We don't have to modify the basic laws of quantum mechanics to describe unstable particles. The full state of the system is includes the state of the decay products, and what you really have is a coupling from one state to another. No imaginary energies are required to describe this, but you do need to include the states of the decay products in your calculation.

This coupling is symmetric (and the total hamiltonian is therefore still hermitian). Still, it is frequently unlikely that the decay products will re-form the original particle because the decay products are usually more than one particle. This means that the entropy of the products is greater than the entropy of the original particle. It is unlikely that this entropy will decrease, so the part of the quantum state that corresponds to the products is in some sense "lost".

Besides the larger state space, the products have less rest mass than the parent particle, which means that for energy to be conserved they have to have more kinetic energy. This makes the product particles fly far away from the location where they formed and from each other; it is unlikely for them to recombine when separated by a great distance.

What Landau is describing is a trick to calculate certain observables without including the dynamics of the decay products. The imaginary part of the Hamiltonian makes the wavefunction decay in a manner similar to a one-way coupling to another state. Since there are many more possible product states, each of them is nearly empty and this is a reasonable approximation.

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Hi Dan, thanks for you answer. If I understand you correctly, a proper treatment of unstable particles requires thermodynamics. If we ignored entropy considerations, Hamiltonian time evolution would be make the decay $A \to B_1\cdots B_N$ just as likely as $B_1\cdots B_N \to A$. Do you happen to know a good reference that does this in detail? –  user22037 Apr 6 '13 at 15:59
    
@user22037: It doesn't require thermodynamics per se. The "entropy" bit is just the size of the state space of the products, and includes the tendency of those products to have lots of kinetic energy that kicks them away from the parent particle. –  Dan Apr 6 '13 at 18:16
    
@user22037: A truly proper treatment of unstable particles requires including state of the products. This is usually very difficult, and it also includes a lot of stuff you don't necessarily care about. The thermodynamic and kinetic explanations are what one would use to justify neglecting this. –  Dan Apr 6 '13 at 18:37
    
Thanks again. So if the interaction describing the transition $A \leftrightarrow B_1 \cdots B_n$ is local (or short-ranged) in space, then the decay products $B_k$ usually end up too far apart to recombine. I realize this is probably difficult to do in full detail, but is there perhaps a toy model in which it can be done exactly or nearly exactly? That would help me understand. Would appreciate any suggestions. –  user22037 Apr 6 '13 at 19:26
    
@user22037: I don't know of any toy models that can be done by hand. You might want to ask that as another question. –  Dan Apr 6 '13 at 19:46

I think it may also be fruitful to not think of the Hamiltonian as the energy. The Hamiltonian is the generator of time translations, and so an eigenvalue of the Hamiltonian that is complex tells you that there is some decay, as was pointed out by Edoot. This is an important distinction. When we compute the "energy spectrum" of a Hamiltonian, what we are really computing is the frequency spectrum of the time evolution of the system.

One of the better discussions I've seen on HOW you obtain these effective Hamiltonians by directly computing them is in Wen's Quantum Field Theory of Many-Body Systems. In one of the early chapters, he talks about a "quantum RLC circuit". The book is hit or miss, but I think this discussion is clear and a good example of effective field theories.

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Hi webb, thanks for your answer. I see I should have been careful in how I phrased my question: what was really bothering me was the fact that the Hamiltonian had complex eigenvalues. I understand "the frequency spectrum of the time evolution," but I don't understand how to reconcile complex frequencies with the requirement that $H$ be Hermitian. Thanks for the reference, I'll take a look if it ever gets returned to my university library. –  user22037 Apr 6 '13 at 16:05
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Effective field theories sacrifice hermiticity to get rid of a degree of freedom. This is like adding a damping term to Hamilton's equations in classical mechanics violating symplecticity. I wish I had a good example of that, but I don't. –  webb Apr 7 '13 at 4:17
    
Very interesting analogy, thanks. So if $H$ is non-Hermitian in an effective field theory describing particle decay, are the imaginary parts of the eigenvalues possible outcomes of measurement? Would really appreciate if you know other references that discuss this in detail, seeing as Wen's book is checked out of my university library and I don't know when it's coming back. –  user22037 Apr 7 '13 at 20:10
    
I don't have another reference, unfortunately. Experimentally, the complex part of a resonance in, say, high energy detector measurements, manifests as having a line width. So if you measure some Lorentzian distribution around a resonant peak of $\omega_0$ in some experiment, the width is determined by the decay time. –  webb Apr 21 '13 at 15:57

We don't need modify postulates.

$ E = E_0 - \frac{1}{2}i \Gamma $ only describes partially the system. $x$ particle disappear, and $y$ particle appears (one or more).

Initialy: $E_x = E_0$ and $E_y = 0$

Finally: $E_x = 0$ and $E_y = E_0$

Add $- \frac{1}{2}i \Gamma$ because of we need a decay:

$$ \exp(-iHt) = \exp(-iE_0t)\exp(-\frac{1}{2} \Gamma t) $$

$\exp(-\frac{1}{2} \Gamma t)\rightarrow 0$ fast = particle disappear.

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Hi Edoot, thanks for your answer. I understand that $-\frac{1}{2}i \Gamma$ makes $\psi \to 0$. What I do not understand is how to make sense of this using a Hermitian $H$. The term $-\frac{1}{2}i \Gamma$ makes $H$ non-Hermitian, in violation of the postulates. –  user22037 Apr 6 '13 at 16:09

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