# PT symmetric non Hermitian formulation of QM

1. Nov 27, 2005

### CarlB

I haven't read this yet, but I'm putting it up here for discussion as it seems so fascinating:

PT-Symmetric Versus Hermitian Formulations of Quantum Mechanics
Carl M. Bender, Jun-Hua Chen, Kimball A. Milton
A non-Hermitian Hamiltonian that has an unbroken PT symmetry can be converted by means of a similarity transformation to a physically equivalent Hermitian Hamiltonian. This raises the following question: In which form of the quantum theory, the non-Hermitian or the Hermitian one, is it easier to perform calculations? This paper compares both forms of a non-Hermitian $ix^3$ quantum-mechanical Hamiltonian and demonstrates that it is much harder to perform calculations in the Hermitian theory because the perturbation series for the Hermitian Hamiltonian is constructed from divergent Feynman graphs. For the Hermitian version of the theory, dimensional continuation is used to regulate the divergent graphs that contribute to the ground-state energy and the one-point Green's function. The results that are obtained are identical to those found much more simply and without divergences in the non-Hermitian PT-symmetric Hamiltonian. The $\mathcal{O}(g^4)$ contribution to the ground-state energy of the Hermitian version of the theory involves graphs with overlapping divergences, and these graphs are extremely difficult to regulate. In contrast, the graphs for the non-Hermitian version of the theory are finite to all orders and they are very easy to evaluate.
http://www.arxiv.org/abs/hep-th/0511229

 At a quick glance, it appears to be a more natural way of including a Wick rotation. My hope would be that something like this would be a particularly natural way of implementing a theory where proper time is promoted to be an element of the geometry of space time, as I've proposed in the past. Along that line, it's interesting that, it's interesting that Kimball is a follower of Schwinger in Schwinger's later years which rejected the existence of the vacuum, which I also agree with. For more on this see:

http://www.arxiv.org/abs/hep-th/9901011
http://www.arxiv.org/abs/hep-th/9811054
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Carl

Last edited: Nov 27, 2005
2. Dec 19, 2005

### lynxman72

I'm actually an undergrad where Bender teaches and I've gotten to know him a little bit, he does a lot of work in this area, I think it's pretty much all he works on these days (that and coaching the Putnam team!), so you may want to check out his other recent publications as well:

http://wuphys.wustl.edu/~cmb/cmbpubs.html

3. Dec 19, 2005

### CarlB

Thanks for reminding me of this. I was going to get back to it, and it seems like my brain must have been working on it in the background.

It seems like the heart of the theory goes back to this paper:

Complex Extension of Quantum Mechanics
Carl M. Bender, Dorje C. Brody, Hugh F. Jones
Abstract: It is shown that the standard formulation of quantum mechanics in terms of Hermitian Hamiltonians is overly restrictive. A consistent physical theory of quantum mechanics can be built on a complex Hamiltonian that is not Hermitian but satisfies the less restrictive and more physical condition of space-time reflection symmetry (PT symmetry). Thus, there are infinitely many new Hamiltonians that one can construct to explain experimental data. One might expect that a quantum theory based on a non-Hermitian Hamiltonian would violate unitarity. However, if PT symmetry is not spontaneously broken, it is possible to construct a previously unnoticed physical symmetry C of the Hamiltonian. Using C, an inner product is constructed whose associated norm is positive definite. This construction is completely general and works for any PT-symmetric Hamiltonian. Observables exhibit CPT symmetry, and the dynamics is governed by unitary time evolution. This work is not in conflict with conventional quantum mechanics but is rather a complex generalisation of it.
http://www.arxiv.org/abs/quant-ph/0208076

I think that this sort of thing is the right way to push physics. It is a fact of symmetry that when any sort of objects condense together to form a new phase, there is only limited similarity between the symmetry of the unbound objects and the symmetry of the condensation. I wonder if this is what he's thinking of too. For example, see:
Bound States of Non-Hermitian Quantum Field Theories
http://www.arxiv.org/abs/hep-th/0108057

Since our whole experience of physics is a hierarchy of condensations (quarks form to form nuclei, nuclei and electrons together form atoms, etc.), one should not expect that the fundamental theory underlying it all is restricted by the same symmetries that are apparent in the standard model (which pretty much everyone agrees is only an "effective" theory).

One of the reasons I'm interested in all this is that I'm playing around in Clifford algebras with a single hidden dimension, we can call it "s". A Clifford algebra is similar to the Dirac matrices in that they satisfy a set of anticommutation relations that are pretty much the same, only, in this case, with one extra spatial dimension. For example,

$$\hat{x}\hat{x} = 1$$
$$\hat{y}\hat{y} = 1$$
$$\hat{z}\hat{z} = 1$$
$$\hat{s}\hat{s} = 1$$
$$\hat{t}\hat{t} = -1$$

and any two elements anticommute.

With this Clifford algebra, which is called CL(4,1), the product of the generators is an element of the algebra that makes a natural imaginary unit i. That is, $$\hat{x}\hat{y}\hat{z}\hat{s}\hat{t}$$ squares to -1 and commutes with everything in the algebra.

Now in this algebra, the natural time operator is $$\hat{t}$$ and the natural parity operator is $$\hat{x}\hat{y}\hat{z}\hat{s}$$, and the product of these two just happens to be the natural imaginary unit. So making the Hamiltonian PT symmetric in the usual physics is equivalent, in the Clifford algebra CL(4,1) to making the Hamiltonian Hermitian.

Was this understandable? I've just downloaded from Arxiv all his recent articles. After I've read them, I'll probably write a letter. I just posted a note on the Euclidean Relativity website.

Euclidean Relativity uses a hidden dimension. A recent paper is that of JMC Montanus in the September 2005 issue of Foundations of Physics. The article is accessible at an undergraduate level and will surprise and shock you. If the local physics library at your campus doesn't have a hard copy, then it will be available free for downloading because Found of Phys is a peer-reviewed Springer-Verlag publication.

The reason for the use of Clifford algebra in uniting gravitation with QM is best shown by the article:

A. Lasenby, C. Doran, & S. Gull, "Gravity, gauge theories and geometric algebra," Phil. Trans. R. Lond. A 356: 487-582 (1998),

which is also available at your library. Reading the above will be a real shock if you understand gravitation, but otherwise may be a bit much.

Carl

I realized I need to clarify what I mean by writing the Clifford algebraic constants as P and T.

The Clifford algebaic $$p = \widehat{xyzs}$$ and $$t = \hat{t}$$ are related to the usual P and T operators but are not identical. The usual P and T operators change the parity or reverse time of complete states. The Clifford algebraic operators are different first in that they are only "intrinsic" operators that do not change the position or time dependency of a wave function. That is, the Clifford algebraic operators are applied only to the wave function and this is something that won't make a lot of sense except in theories that put the elementary particles into a Clifford algebra such as my own or Trayling's:
http://www.arxiv.org/abs/hep-th/9912231

The second difference is that the usual P and T operators also change the intrinsic properties of the wave function they are applied to while the Clifford algebraic operators are used in eigenvector and eigenvalue relations in a manner similar to how the spin operators are used.

For example, consider the action of the usual P operator on two states of opposite parity:

$$P |+> = e^{i\alpha}|->$$
$$P |-> = e^{i\alpha'}|+>$$

where alpha is some phase angle. In contrast, the Clifford algebra operator equations would be:

$$p |+> = +1\;|+> = |+>$$
$$p |-> = -1\;|-> = -|->$$

This alternative definition of the P and T operators is uncommon, but is given in the relativistic quantum mechanics book by Landau and Lipsh_itz.

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Last edited: Dec 20, 2005
4. Dec 20, 2005

### lynxman72

Carl, I'm sorry, I'll get back to you soon on it, right now just trying to finish up finals (last day is tomorrow)...I'll also check out that paper you referenced that you said should be accessible, we'll see what i can do with it, as this is actually my first semester studying math and physics (I don't mean formally studying them, just studying them period)...anyways, talk to ya soon

5. Oct 24, 2007

### arige

Hi,
Why the eigenvalues of PT operator can be set to a unit?