# Hamiltonian with complex potential

eljose
let,s suppose we have the Hamiltonian H=T+V but V is V=F(z) being z=a+ix then would the energies be real?..thanks.

## Answers and Replies

Observable
clearly not.

Broken
Imaginary components to the potential can be useful, though. If you look at the time evolution of the wavefunction, you'll notice that the probability is no longer conserved, which can be used to describe a particle in a decaying state, i.e. leaving the system described by your hamiltonian.

inha
That's commonly done via $$V=V_0+i\Gamma$$ instead of a potential whose argument is complex. I don't see that being a very physical scenario either unless we're talking about an oscillatory potential.

Science Advisor
Yea this is done all the time for instance in nuclear physics. It is known as the optical model.

Science Advisor
Homework Helper
eljose said:
let,s suppose we have the Hamiltonian H=T+V but V is V=F(z) being z=a+ix then would the energies be real?..thanks.

Assuming that the kinetic energy operator $$\hat{T}$$ is selfadjoint for a certain domain (of course, one could use the Schroedinger picture and the analysis of selfadjointness would be much easier), one has to have that, assuming boundness for both operators entering the Hamiltonian,
$$\hat{V} =\hat{V}^{\dagger}$$

It's all simple, if one uses the Schroedinger picture. Then the operatorial equality induces certain requirements on the V(z).

Daniel.

pmb_phy
eljose said:
let,s suppose we have the Hamiltonian H=T+V but V is V=F(z) being z=a+ix then would the energies be real?..thanks.
Maybe and maybe not. For examle; if F(z) = a2 + x2 then the energy must be real since F(z) is real. But for the life of me I can't see how this would correspond to something in the real world. This would imply that there are complex observables. However the Hamiltonian (in position space and you've indicated) must correspond to an operator which has real eigenvalues. If it doesn't then it violates one of the axioms of QM since the Hamiltonian would then not be Hermitian.

Pete

See the above post about decaying states.

QMrocks
we use complex potential to represent system that leaks to the outside, its called 'coupling function'.

pmb_phy
QMrocks said:
we use complex potential to represent system that leaks to the outside, its called 'coupling function'.
Let me get back on this. Someone explained what was meant here but I need a solid reference to go through, e.g. Cohen-Tannoudji, Sakuri, etc.

Pete

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Staff Emeritus
Science Advisor
Gold Member
Complex potentials are used to crudely model particle decays in nonrelativistic quantum mechanics.

Consider the nonrelativistic Hamiltonian $H_0=\frac{p^2}{2m}+V_0$. The wavefunctions for this Hamiltonian are of the form $\psi (0) exp(\frac{-iE_0t}{\hbar})$. Since $H_0$ is Hermitian, the time evolution operator is unitary, and probability is conserved. That is, $|\psi (0)|^2=|\psi (t)|^2$ for all $t$.

Now make the following substitution:

$V_0 \longrightarrow V=V_0-\frac{1}{2}i\Gamma$,

where $\Gamma$ is a positive real number.

What happens? The new eigenvalues of the Hamiltonian are changed as follows:

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

Now the Hamiltonian is not Hermitian, and consequently the time evolution operator is no longer unitary. The new wavefunctions are of the form:

$\psi (t)= \psi (0) exp(\frac{-iE_0t}{\hbar})exp(\frac{-\Gamma t}{2\hbar})$,

and the probability density is now given by

$|\psi (t)|^2 =|\psi (0)|^2exp(\frac{-\Gamma t}{\hbar})$.

A reference on this is Subatomic Physics, 2ed, Section 5.7, by Frauenfelder and Henley. There's more to their discussion than what I have presented here, but you get the idea.

lalbatros
Hello

Non-Hermitian Hamiltonian are indeed useful to study decaying systems.
A long time ago, I computed the time-dependent Stark effect on hydrogen (influence of an electric field). This leads -of course- to quantum oscillations, observable from their emissions in beam-foil experiments. The decay times are also depending on the electric fields. The easiest way to modelize this system is by using the standard Hamiltonian for hydrogen with a non-hermitian perturbation term leading to a line-width. The use of the density matrix is quite natural since the hydrogen beams are decaying.

There are many ways to introduce a non-hermitian term. Like for example some approximation for the interaction with quantized fields. The simplest way is by adding a decay term as explained by Tom. I don't directly see the physical meaning that an imaginary space coordinate could have, but I would suggest to try some insight by a Taylor series development and to see how a link could be found with impulsion (Tom's procedure shows a link with energy). A potential gradient term will appear, which is a force, and this "force" wil be the responsible for the decay.

Another point for reflexion is how the complex potential is defined based on its real-valued counter-part: this is continuation.

kamalmgu
Hamilton And Lagrangean Mechanism

what are the advantages of hamiltonian mechanisms over lagrangian mechanics while dealing with quantum field theory

kamalmgu
scattering theory

while performing typical scattering experiment if the depth of the potential is large the condition is no scattering that the depth is about 180 degree the condition is called Ramswood - thowsend effect it can also take place while we are using complex potential

lalbatros
kamalmgu,

I googled for "Ramswood - thowsend effect" but could not find anything.
Could you explain a little more, or point me to a reference (web preferably).

Thanks,

Michel

da_willem
A non-Hermitian term in the Hamiltonian is also used in the study of non-equilibrium properties of single molecules by means of the Non-Equilibrium Green's Function (NEGF) method. Here so-called self-energy matrices are added to the molecular Hamiltonian to take into account the effect of (macroscopic) contacts without having to consider the huge amount of particles in the contacts.

As said before, this addition of a non-Hermitian term makes that the number of particles is not necissarily conserved. Also mentioned is the finite lifitime of the previously stationary states upon the addition of a non-Hermitian term. Accompanying this, by the uncertainty principle, is a broadening of the eigenergies. Just as you would expect for a molecule attached to contacts with overlapping MOs etc...

pmb_phy
kamalmgu said:
what are the advantages of hamiltonian mechanisms over lagrangian mechanics while dealing with quantum field theory
Its part of the axioms for one thing since an axiom is stated in terms of a Hamiltonian.

The Lagrangian is expressed, in part, in terms of velocity for which there is never such an operator even defined. The Hamiltonian is expressed, in part, in terms of canonical momentum for which there is always an operator defined.

Pete

eljose
my question is could an energy be real?..if this happened with E_{n}=E*_{n} ten we would have with b(x) the complex part of the potential:

$$\int_{-\infty}^{\infty}dx(|\phi_{n}|^2)b(x)=0$$

but if the potential is complex then b is different from 0.

beautiful1
eljose said:
let,s suppose we have the Hamiltonian H=T+V but V is V=F(z) being z=a+ix then would the energies be real?..thanks.

Hi eljose,
As I understand, if F(z) is real, i.e. if H is real, the eigenergies will be real. And I think it is okay if z is a complex variable. Or even an operator with real and complex parts. As an example, you might consider the ladder operator a=x-ip, or its expectation value <a>=<x>-i<p>, where x and p are position and momentum of an oscillator (or in-phase and in-quadrature components of an electric field).

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Careful
eljose said:
my question is could an energy be real?..if this happened with E_{n}=E*_{n} ten we would have with b(x) the complex part of the potential:
$$\int_{-\infty}^{\infty}dx(|\phi_{n}|^2)b(x)=0$$
but if the potential is complex then b is different from 0.
Hi, let's restrict to one dimension, then basically you have to solve the problem
E f(x) = -f''(x) + ixf(x).
make an ansatz f(x) = e^{-x^2} g(x).
The equation reduces to Eg(x) = - g''(x) + 4xg'(x) - 4x^2 g(x) + ixg(x) and see how far you get by making an analytical anszatz for for g(x). Even though this dynamics is non unitary it is still possible it has a stationary stable state of constant energy.