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Is the Schrödinger equation still valid if we use a non-Hermitian Hamiltonian with it? By this I mean does:

$$\hat{H}\psi(t) = i\hbar\frac{\partial}{\partial t}\psi(t)$$

if $\hat{H}$ is not Hermitian?

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    $\begingroup$ "valid" in what sense? $\endgroup$ – ACuriousMind Feb 28 '17 at 16:47
  • $\begingroup$ I mean, if $\hat{H}$ is non Hermitian ( of course have complex eigenvalues ) and applied to an eigenfunction $\psi(t)$, do that equal to $i\hslash \frac{\partial}{\partial t}\psi(t)$ or something else? $\endgroup$ – Ahmed M. Farouk Feb 28 '17 at 16:59
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    $\begingroup$ Required reading: Bender 2007. $\endgroup$ – Cosmas Zachos Feb 28 '17 at 20:21
  • $\begingroup$ Related: physics.stackexchange.com/q/16678 $\endgroup$ – Emilio Pisanty Mar 1 '17 at 15:53
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If you have some arbitrary linear operator $\hat A$, there's nothing stopping you from formulating the differential equation $$ i\partial_t \Psi = \hat A \Psi, $$ but you also have no guarantee that its solutions will play nicely or even exist.

In the simplest case, you can take $\hat A=-ia\mathbb I$, and your Schrödinger equation reads $\partial_t \Psi = -a\Psi$, giving you exponential decays of the form $\Psi(t) = e^{-at}\Psi(0)$. If you have a postive real part, as in e.g. $\hat A=+ia\mathbb I$, then you'll have an exponential growth as $\Psi(t) = e^{+at}\Psi(0)$, which isn't terrible. You can also have mixtures between these, such as e.g. a two-by-two matrix $$ A=\begin{pmatrix} ia&0\\0&ib\end{pmatrix}, $$ and you'll get different decay constants for the different coordinates. From here it's easy to extend to arbitrary finite complex matrices, where the solutions will obviously be a bit more complex. However, you need to be careful, because if you break the premise of hermiticity you also lose the guarantee that your operator will be diagonalizable, such as a Jordan block of the form $$ A=\begin{pmatrix} ia&1\\0&ia\end{pmatrix}, $$ which cannot be reduced further; as such, you probably want to demand that your operator be normal, or some similar guarantee of niceness.

You should be double wary, of course, in infinite dimensions, particularly if there is the chance that $\hat A$ will have a definite spectrum whose imaginary part is unbounded from above. A simple example of that type is $$ i\partial_t \Psi(x,t) = ix\Psi(x,t), $$ which is just about solvable as $\Psi(x,t) = e^{xt}\Psi(x,0)$, but here if your initial condition is at large positive $x$ the rate of growth becomes unbounded. From there, it isn't hard to envision the possibility that with slightly more pathological operators you could completely lose the existence of the solutions.


That said, non-hermitian hamiltonians are used with some frequency in the literature, particularly if you're dealing with resonances in a continuum or decaying states. The book referenced here might be a good starting point if you want to read about those kinds of methods. At a more gritty level, you can ask Google Scholar and it will yield ~20k results, with many of those related to something called PT-symmetric quantum mechanics, but that's probably rather more information than what you're after at the moment.

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There is nothing stopping you from writing $\hat H\Psi(x,t)=i\hbar \partial_t\Psi(x,t)$ for arbitrary $\hat H$. The physics is in $\hat H$, not in the differential equation.

Now, if you want to get there, why not write even more generally $$ \hat{\cal O}\Psi(x,t)=\partial_t\Psi(x,t) $$ and get rid of the $i\hbar$ factor?

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Let's suppose that we have an operator $\hat{\mathcal{O}}$ such that $i \hat{\mathcal{O}} |\psi\rangle = \partial_t |\psi\rangle$. The question is then: What is $\partial_t (\langle \psi | \psi \rangle)$? Well, by definition of the adjoint we have $\partial_t \langle \psi | = \langle \psi| (i \hat{\mathcal{O}})^\dagger = -i \langle \psi| \hat{\mathcal{O}}{}^\dagger$, and so we have $$ \partial_t \left(\langle \psi | \psi \rangle \right) = \left( \partial_t \langle \psi | \right) | \psi \rangle + \langle \psi |\left( \partial_t | \psi \rangle \right) = -i \langle \psi |\hat{\mathcal{O}}{}^\dagger| \psi \rangle + i \langle \psi |\hat{\mathcal{O}}| \psi \rangle = i \langle \psi |(\hat{\mathcal{O}} - \hat{\mathcal{O}}{}^\dagger)| \psi \rangle $$

This means that if $\hat{\mathcal{O}}$ is Hermitian, then $\partial_t (\langle \psi | \psi \rangle) = 0$ for all states (and vice versa). But under the usual interpretation of QM, $\langle \psi | \psi \rangle$ is the norm of the state $ | \psi \rangle$, and so we would like it to be constant (and usually equal to 1.) The fancy words for this are that a Hermitian Hamiltonian implies unitary time evolution.

There are some cases where we don't actually want unitary time evolution; the most well-known (which was mentioned by Emilio Pisanty in his answer) are decay processes, which can be modeled quite nicely by a non-Hermitian Hamiltonian. In this case, you want the probability of the particle's existence to decrease (exponentially) with time, so the non-unitarity of the time evolution is actually desirable.

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Someone else has already mentioned PT-symmetric Quantum Mechanics. To expand on that, and what I believe is the spirit of your question rather than the letter, sure; you can write down the Schrödinger equation using a non-Hermitian Hamiltonian. The interesting question from a physical perspective is whether this describes something physical - i.e. whether it describes something we can measure in the lab, that we can understand to fit within the framework of QM.

Introductory texts (like Griffiths) often suggest that we require observables to be represented by Hermitian operators because their eigenvalues (aka the results of measurements we can make in a lab) will be real. In this line, PT-symmetric QM has been suggested as an extension of QM, where the Hamiltonian is now required to have Parity and Time symmetry: This means that the Hamiltonian commutes with the product of the parity and time reversal operators, $[\mathcal{\hat{P}\hat{T}},\hat{H}] = 0$

It can be shown that Hamiltonians that satisfy this property have real eigenvalues (i.e. possible measurements of the energy will always yield a real number). There are additional issues related to unitary time evolution, as mentioned in another answer. Most physicist's would see non-unitarity as at least a big a problem as non-real energy values. This issue can allegedly also be solved.

I don't know how well motivated this is. I gained the impression from attending a (single) talk that people in the business of dealing with issues related to the foundations of Quantum Mechanics hope that such a generalisation may be meaningful. Until there is some insight into the empirical status of PT-symmetric QM this is debatable, IMO.

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