Is every Hamiltonian necessarily Hermitean?

[AxiomOfChoice]

I know the eigenvalues of a Hermitean operator are necessarily real, and we want energies to be real...but isn't it possible for non-Hermitean operators to have real eigenvalues? If that's so, shouldn't it be possible for at least some Hamiltonians to be non-Hermitean?

Also, is it possible for any real differential operator of the form

<br /> - \frac{\hbar^2}{2m} \Delta + V<br />

to be non-Hermitean?

 

[G01]

Consider the following situation:

We have a system in which energy is conserved, so that it is time translation invariant:

Thus if we translate it forward in time, by t seconds, and then backward by -t, the system must return to it's initial configuration. In other words, the time translation operator must be unitary:

U^{\dagger}U=I

But time translation can be expressed in terms of the Hamiltonian operator:

For infinitesimal translations:

U=(I+iH\delta t)

From here, you can explicitly multiply out U^{\dagger}U. Drop terms of order \delta t^2, since it is an infinitesimal quantity.

You should then see that for U to be unitary,

H^{\dagger}=H

Thus, the moral is that non hermitian H would imply that energy is not conserved, since we would not have time translation invariance.

 

[arkajad]

Take for instance

V=\begin{pmatrix}0&amp;f\\0&amp;0\end{pmatrix}

Its eigenvalues (0) are real. It is non-Hermitian. You can easily exponentiate

e^{itV}=I+itV=\begin{pmatrix}1&amp;itf\\0&amp;1\end{pmatrix}

Does not look like a very good candidate for a unitary evolution...

On the other hand it can be used for an evolution of an open quantum system (as an extra term to the Hamiltonian evolution) as follows:

\dot{\rho}=-i[H,\rho]+V\rho V^\dagger -\frac12\{V^\dagger V,\rho\}

which is trace and positivity preserving. But that's a different story, with a dissipation and wave function collapses.

 

[mike372]

It seems there do exist non-hermitian hamiltonians with real eigenvalues. These are known as PT-symmetric hamiltonians and have been developed by Carl Bender, see these reviews:

arXiv:hep-th/0703096v1
arXiv:0810.5643v3

 

[DrFaustus]

AxiomOfChoice -> The reason why you *want* a self-adjoint Hamiltonian (and not only Hermitian...) is that for s.a. operators you can spectrally decomose them and define functions of those operators, not so much that the eigenvalues are real. For instance, say the eigenvalues of some operator are complex, then it could be that the physical value is only encoded in the real part of that number. Or maybe you'd have to consider the imaginary part, or its modulus. In other words, there are various ways to extract a real number from a complex one. But for an operator that is not self-adjoint, you would not be able to define, say, its exponential and hence the time evolution of the state.

If an operator is or is not hermitian/self-adjoint/whatnot is not for us to argue about but is a property of the operator itself. And you will have to specify the operator explicitly together with its domain of definition. Only then can you establish some of its properties.

GO1 -> Your argument is either not really correct. Your first requirement, that you must get back to your original point after going forward in time by t and then reversing, does not imply that U must be unitary. What it says is only that U(-t)U(t) = 1, and not that U^{-1} = U^*, which is the requirement for unitarity.

Then you write the infinitesimal form of U, but it's not clear where is that coming from. Why does it have that form? Because the full form is the usual exponential? But can you define that exponential? Also, for a system with an external varying field, like a background changing electric field, energy is *not* conserved. This is reflected by a Hamiltonian that depends on time. Yet it will still be self-adjoint.

mike372 -> I had a quick look at the above papers, and it would seem that nowhere is time evolution of a state or of an observable defined. Not going to say anything really as I have not studied them and neither am I an expert in functional analysis, but they seem mathematically shaky.

 

[arkajad]

The funny thing is that in QM we are playing with "operators" and the first thing that operators do is that they operate on states. Yet nowhere in the standard QM an operation menaing is given to the "operation of an observable on a state". There is no meaning given to, for instance, operation on a state by the energy or position operator. Instead, meaning is given to the operation of its exponential or to its eigenstates, eigenvalues, expectation values.

And for just "operating on states" operators do not have to be selfadjoint. On the other hand an interpretation is given to operations by some non-selfajoint opeartors like creation and annihilation operators.

 

[dpackard]

The funny thing is that in QM we are playing with "operators" and the first thing that operators do is that they operate on states. Yet nowhere in the standard QM an operation menaing is given to the "operation of an observable on a state". There is no meaning given to, for instance, operation on a state by the energy or position operator. Instead, meaning is given to the operation of its exponential or to its eigenstates, eigenvalues, expectation values.

And for just "operating on states" operators do not have to be selfadjoint. On the other hand an interpretation is given to operations by some non-selfajoint opeartors like creation and annihilation operators.

I don't know about that. The operator of an observable operates on states by returning the possible values one can measure. One could just as well say that there is no "meaning" in saying we have rotated or time translated a state except in the context of what is being measured. So your gripe is misplaced IMO.

 

[arkajad]

The operator of an observable operates on states by returning the possible values one can measure.

The operator operates on states by returning states and not "values". Values can be calculated one way or another. But what operator A returns when acting on state x is another state, namely Ax.

 

[dpackard]

Yes, mathematically that is obviously true. I was speaking imprecisely, forgive me. But, to belabor the point, translation operators don't actually "translate" states either - they map vectors that we interpret as being at position to vectors at another. Observables map states that are superpositions of eigenkets to another superposition where each eigenket is weighted by its measurement value. There is plenty of "meaning" to extract there.

However, what IS different is that observable operators don't represent any sort of physical action, such as rotations or translations, but they are hardly unique in this regard. In fact, I'd argue rotations and space translations are never used in an "action" way either, only mathematically - it seems more like that only operators used to represent physical occurrences are projections and time evolution (propagators).