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## Main Question or Discussion Point

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.

- Thread starter eljose
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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.

- #2

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clearly not.

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- #5

Haelfix

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

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Assuming that the kinetic energy operator [tex] \hat{T} [/tex] 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,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.

[tex] \hat{V} =\hat{V}^{\dagger} [/tex]

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

Daniel.

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Maybe and maybe not. For examle; if F(z) = aeljose said:

Pete

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MalleusScientiarum

See the above post about decaying states.

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- #10

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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.QMrocks said:

Pete

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Tom Mattson

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

Now make the following substitution:

[itex]V_0 \longrightarrow V=V_0-\frac{1}{2}i\Gamma[/itex],

where [itex]\Gamma[/itex] is a positive real number.

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

[itex]E_0 \longrightarrow E=E_0-\frac{1}{2}i\Gamma[/itex].

Now the Hamiltonian is

[itex]\psi (t)= \psi (0) exp(\frac{-iE_0t}{\hbar})exp(\frac{-\Gamma t}{2\hbar})[/itex],

and the probability density is now given by

[itex]|\psi (t)|^2 =|\psi (0)|^2exp(\frac{-\Gamma t}{\hbar})[/itex].

A reference on this is

- #12

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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.

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

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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

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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

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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...

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

The Lagrangian is expressed, in part, in terms of velocity for which there is

Pete

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[tex]\int_{-\infty}^{\infty}dx(|\phi_{n}|^2)b(x)=0 [/tex]

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

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Hi eljose,eljose said:

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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Hi, lets restrict to one dimension, then basically you have to solve the problemeljose said:

[tex]\int_{-\infty}^{\infty}dx(|\phi_{n}|^2)b(x)=0 [/tex]

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

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.

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