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? *

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    $\begingroup$ 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 $\endgroup$
    – Mithrandir24601
    Dec 8, 2018 at 12:51
  • $\begingroup$ @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. $\endgroup$
    – user820789
    Dec 8, 2018 at 19:06
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    $\begingroup$ 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 $\endgroup$
    – Mithrandir24601
    Dec 8, 2018 at 20:02
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    $\begingroup$ 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 $\endgroup$
    – Mithrandir24601
    Dec 9, 2018 at 17:34
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    $\begingroup$ @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) $\endgroup$
    – Mithrandir24601
    Dec 14, 2018 at 16:10

1 Answer 1


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

  • $\begingroup$ In the PT set of equations third line, x is not in the RHS. The free variables on both sides don't match. $\endgroup$
    – AHusain
    Jan 21, 2019 at 13:44
  • $\begingroup$ @AHusain whoops - fixed now, at least down to normal abuses of notation $\endgroup$
    – Mithrandir24601
    Jan 21, 2019 at 14:40

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