# Eigenvalues of CPT operator

Suppose we define an operator CPT that carries out the CPT transformation: $$\text{CPT}|\Psi\rangle = A|\Psi\rangle$$ where A is just a constant. Or put another way, the states of our theory are eigenfunctions of the CPT operator. -Source

What do the eigenvalues of the CPT operator mean? *

• I'm not sure which 'idea' you're referring to here - there's an issue that you're saying e.g. "let $⟨0|$ represent a charge transformation" - this isn't possible as $⟨0|$ is a state, not a transformation. In the meantime, I'll have to put this on hold but I think we can edit this down to the actual question you want answered relatively quickly – Mithrandir24601 Dec 8 '18 at 12:51
• @Mithrandir24601 I have updated the post to reflect the feedback given. Apologies for the lack of clarity of this question.. I was just sketching this image in 3d given the cube configuration above. Then I noticed all the symmetries & translations & it kind of blew my mind! Maybe it's such an obvious thing & I had just not seen it before.. I however still find it interesting that given this setup, I'm not sure it's possible to determine which state was the starting state of the sequence. – meowzz Dec 8 '18 at 19:06
• Well, it's certainly clear now - the previous version still had a question in there as well and I do think that could be a good question, with a bit of work - in the paper you originally mentioned - they used e.g. $\left<t_{01}\right| = \left<0\right| + \left<1\right|$ (up to normalisation, anyway), which would be more like $⟨∅|\left(C+P\right)$, while the 'CP transformation' would be more like $⟨∅|CP$ (although I've skipped a couple of things that would need more than a comment to explain), so I'm still not sure which exactly you meant – Mithrandir24601 Dec 8 '18 at 20:02
• Just a note for anyone wondering about whether or not this in on topic - there's CPT symmetry as referred to in QFT and particle physics (neither of which are explicitly covered by this SE), then heavily related is the CPT symmetry of 'PT-symmetric/non-Hermitian quantum theory', which is more like fundamentals of quantum information theory and (I'd certainly argue) is at least relevant from a computational perspective, although people are of course, free to disagree – Mithrandir24601 Dec 9 '18 at 17:34
• @glS If I'm remembering right, it's not unitary, but it's the product of the charge, parity and time 'reversal' operations (or at the least, something equivalent) – Mithrandir24601 Dec 14 '18 at 16:10

## Background - CPT (and $$\mathcal{CPT}$$) symmetry in QFT

I think the first step in answering is this is explaining what CPT (also, $$\mathcal{CPT}$$) symmetry from a quantum computing perspective is not. If this same question on physics SE was asked, I would refer them to chapters 23 and 40 of Srednicki's Quantum Field Theory for the full-blown explanation. However, while that is useful background for understanding the CPT operation from a computational perspective, the important point is that the Parity $$\left(\mathcal P\right)$$ and Time $$\left(\mathcal T\right)$$ reversal transformations are operations that act on the spatial and temporal co-ordinates, giving unitary Parity $$\left(P\right)$$ and Time $$\left(T\right)$$ operators on the field, as well as additional Charge conjugation operations $$\left(\mathcal C\text{ and }C\right)$$ and so, may change depending on the particle being looked at/used. Implementing these transformations on a quantum computer would then mean that different 'implementations' of computer could act differently under the $$\mathcal C,\, \mathcal P$$ and $$\mathcal T$$ operations, which is not so good.

## $$\mathcal{PT}$$-symmetric Hamiltonians

However, what this does lead to is the realisation that Hamiltonians can be non-Hermitian, yet have real eigenvalues due to having $$\mathcal{PT}$$ symmetry (of the 'reversing the spatial and temporal co-ordinates' sense) as discovered by Bender and Boettcher. This symmetry can be broken (the Hamiltonian has real eigenvalues) or unbroken (the Hamiltonian has complex eigenvalues), with a phase transition (often known as the exceptional point) in between. This allows for the definition of $$\mathcal P\text{ and }\mathcal T$$ ($$\mathcal T$$ is complex conjugation), as well as the $$\mathcal{PT}$$ inner product: for a state $$\psi\left(x\right)$$ and Hamiltonian $$H$$ with eigenstates $$\phi_n\left(x\right)$$,

\begin{align} \mathcal P\left(x, y\right) &= \delta\left(x+y\right) \\ &= \sum_n\left(-1\right)^n\phi_n\left(x\right)\phi_n\left(-y\right) \\ \mathcal{PT}\psi\left(x\right) &= \psi^{\mathcal{PT}}\left(x\right) \\ &= \int dy\, \mathcal P\left(x, y\right)\mathcal T \psi\left(y\right) = \psi^*\left(-x\right) \\ \left\langle\psi|\chi\right\rangle^{\mathcal{PT}} &= \int dx\,\psi^{\mathcal{PT}}\left(x\right)\chi\left(x\right) \end{align}

As per Bender, Brody and Jones, in the unbroken regime, an operator $$\mathcal C$$ (resembling the charge conjugation operator in QFT mentioned above) can be defined, as well as the $$\mathcal{CPT}$$ inner product $$\left\langle\psi|\chi\right\rangle^{\mathcal{CPT}}$$:

## Eigenvectors of the $$\mathcal{CPT}$$ operator

In the unbroken phase, the eigenvalues of the Hamiltonian are real. As the parity reversal operator has no effect on a number and the time reversal operator is complex conjugation, we have that $$\mathcal{PT}E_n\phi_n = E_n\mathcal{PT}\phi_n$$ and so, the eigenstates of the Hamiltonian are simultaneously eigenstates of the $$\mathcal{PT}$$ operator (it's assumed that the eigenvectors are non-degenerate for simplicity, although this breaks down at the exceptional point).

As shown by Weigert, we also have the completeness relation $$\sum_n\left(-1\right)^n\phi_n\left(x\right)\phi_n\left(y\right) = \delta\left(x-y\right)$$.

We also have that $$\left(\mathcal{PT}\right)^2\phi_n\left(x\right) = \mathcal{PT}\phi^*_n\left(-x\right) = \phi_n\left(x\right)$$, giving eigenvalues $$e^{-i\omega_n}$$ as $$\mathcal Te^{-i\omega_n} = e^{i\omega_n}\mathcal T$$. This eigenvalue is absorbed into the definition of the eigenvector so that $$\phi_n\left(x\right)\rightarrow e^{-\frac{1}{2}i\omega_n}\phi_n\left(x\right)$$ and now $$\mathcal{PT}\phi_n\left(x\right) = \phi_n\left(x\right)$$.

This gives that $$\left\langle\phi_n|\phi_m\right\rangle^{\mathcal{PT}} = \int dx\,\phi_n\left(x\right)\phi_m\left(x\right)$$. However, this is not necesarily positive, so we define an additional operator, $$\mathcal C$$, alongside the $$\mathcal{CPT}$$ inner product, in order to satisfy the completeness relation:

\begin{align}\mathcal C\left(x, y\right) &= \sum_{n=0}^\infty \phi_n\left(x\right)\phi_n\left(y\right) \\ \mathcal C\phi_n\left(x\right) &= \left(-1\right)^n\phi_n\left(x\right) \\ \psi^{\mathcal{CPT}}\left(x\right) &= \int dy\, \mathcal C\left(x, y\right)\psi^*\left(-y\right) \\ \left\langle\psi|\chi\right\rangle^{\mathcal{CPT}} &= \int dx\,\psi^{\mathcal{CPT}}\left(x\right)\chi\left(x\right) \end{align}

Computationally, discrete Hamiltonians/systems are more commonly used than continuous ones and helpfully, the above applies in the same/equivalent way to continuous systems (swapping continuous for discrete where applicable) as above.

## The eigenvalues of the $$\mathcal{CPT}$$ operator

Having defined parity and time reversal operators and a new operator that looks like charge conjugation in order to satisfy the completeness relation, we have the answer of what the eigenvalue of the $$\mathcal{CPT}$$ operator is already written down: $$\mathcal{CPT}\phi_n\left(x\right) = \left(-1\right)^n\phi_n\left(x\right),$$ where $$\phi_n$$ is an eigenstate of both the $$\mathcal{CPT}$$ operator and some $$\mathcal{PT}$$-symmetric Hamiltonian.

But what do these eigenvalues mean?

• As per the linked answer, it's analogous to the result in QFT where the eigenstates of the system are invariant under $$\mathcal{CPT}$$ reversal

• The Hamiltonian is in the unbroken regime (that is, it has real eigenvalues)

• As such, the eigenbasis is complete

• the eigenvectors corresponding to the $$-1$$ eigenvalues have a negative $$\mathcal{PT}$$ inner product

• Which feels similar to the notions of positive and negative charge of e.g. positrons and electrons, only, it's not that

• Not much else, really

• In the PT set of equations third line, x is not in the RHS. The free variables on both sides don't match. – AHusain Jan 21 at 13:44
• @AHusain whoops - fixed now, at least down to normal abuses of notation – Mithrandir24601 Jan 21 at 14:40