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 Dec 8, 2018 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. Dec 8, 2018 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 Dec 8, 2018 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 Dec 9, 2018 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) Dec 14, 2018 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. Jan 21, 2019 at 13:44
• @AHusain whoops - fixed now, at least down to normal abuses of notation Jan 21, 2019 at 14:40