This may be really basic but I'm having trouble connecting the following issues:

1) The 2-norm for state vectors is preserved during time evolution

2) The Hamiltonian is a Hermitian operator. With the Schrodinger equation, we can then understand how time evolution acts on a state.

As a consequence of 1) or 2), we can show that time evolution is unitary. But I do not see why 1) or 2) must be true to begin with. Moreover, if only one of them is true, then do I hit some kind of inconsistency?

Let's start from 1) - When we consider time evolution of states, why is there a requirement that inner products be preserved? Obviously, I don't want to assume that evolution is unitary since that's circular reasoning. What does it matter if I fail to preserve the 2-norm of some arbitrary pair of states as they both undergo time evolution?

As for 2), this answer (https://physics.stackexchange.com/a/264439/52363) was quite englightening. If operators do not have to be Hermitian, then neither does the Hamiltonian. As a result, I can get some non-unitary evolution which, of course, does not preserve the 2-norm.

I think I'm a bit muddled with how these concepts come together so any help is greatly appreciated! Which assumptions come first and why do we end up with the quantum theory that we have today?

  • $\begingroup$ 1) needs to hold since otherwise the probability to measure "whatever" would differ from 1. $\endgroup$ – Photon Nov 24 '17 at 10:10
  • $\begingroup$ @Photon, could you make that statement more precise? Are we talking about $\langle \psi\vert\psi \rangle$ or between any arbitrary states i.e. $\langle \phi\vert\psi \rangle$ $\endgroup$ – user1936752 Nov 24 '17 at 10:13
  • $\begingroup$ We are talking about the (squared) norm, not the scalar product: $$\langle \psi|\psi\rangle = \int \psi^\star(x)\psi(x)d^3x = \int |\psi(x)|^2d^3x=\int \rho(x)d^3x$$ where $\rho$ is the probability density. The integral over the whole space of $\rho$ should equal 1. $\endgroup$ – Photon Nov 24 '17 at 10:18
  • 1
    $\begingroup$ Which assumptions come "first" depends on how you want to do quantum mechanics. When you have a theory with a bunch of statements that can all be derived from each other given one of them, which one of them comes "first" is a rather meaningless question - it's a question of taste which of these statement you can motivate best to accept as an axiom, but there's nothing intrinsic to the theory that would make that better or worse than any other choice. $\endgroup$ – ACuriousMind Nov 24 '17 at 11:25
  • $\begingroup$ @ACuriousMind, fair enough. In this case, does it mean both 1) and 2) are independent assumptions? And if I assume 1) but not 2), will I still get unitary evolution or something else? $\endgroup$ – user1936752 Nov 24 '17 at 17:10

Most of the question has already been answered in the comments. I'll summarise and expand on some of the points.

The OP asks why we have a requirement that the 2-norm of vectors be preserved under time evolution. The answer, already given in the comments, is as follows. Let $U(t)$ be the time evolution operator, and $\lvert \psi(t) \rangle$ the state of the system at time $t$. Then $\lvert \psi(t) \rangle = U(t)\lvert \psi(0) \rangle$ where $\lvert \psi(0) \rangle$ is the initial state. We insist that the wavefunction is normalised, which in the usual interpretation means that probabilities "add up to 1". So if $\langle \psi(0)\lvert \psi(0) \rangle=1$. We therefore require that $U(t)$ is unitary to ensure that the wavefunction is then normalised at all times.

If time evolution wasn't unitary, then you may have $\langle \psi(t)\lvert \psi(t) \rangle \neq 1$. If it's greater than one, then it's nonsense (as long as you believe that the wavefunction can be interpreted as a probability amplitude). If it's less than one, then your system is "leaking" information. This is also nonsense in the usual interpretation. Some people use this to model dissipative effects by using an imaginary potential, which translates to non hermitian hamiltonian and therefore (if you believe Schrodinger) non unitary evolution.

As the question you linked points out correctly, we need operators corresponding to observables to be normal. This is because by the spectral theorem, this means that they are diagonalisable by an orthogonal set, which is something we need if measurements are to make sense. Notice however that we can build an hermitian Hamiltonian from non-normal operators. The typical example is the harmonic oscillator, for which $H=a^\dagger a$, but $a$ is not normal. If we require the eigenvalues of a normal operator to be real, then the operator is hermitian. Why do we do this? It's what we are used to. Position, momentum and energy eigenvalues have a simple interpretation if they are real. What if they were complex? This happens when you do scattering, for which the wavefunction is not normalisable, and you may get complex momentum and energies, which in that context may be interpreted as decay times (see these notes, page 312 onwards). However it's not clear how to interpret them in general.

We can summarise the logical chain:

$$\textrm{wavefunctions are probabilty amplitudes} \implies \textrm{time evolution should be unitary}$$

$$\textrm{observables give real measurements} \implies \textrm{corresponding operators should be hermitian}$$

$$\textrm{Schrodinger equation holds} \Leftrightarrow \textrm{time evolution is given by } e^{-iHt}$$

So to answer your question in the comments, if you assume 1) but not 2) you still get unitary time evolution, however you do not necessarily know its mathematical expression.

Notice however that there are cases in QM when we want to interpret things differently. The most common example is scattering states, whose wavefunctions are not interpreted as probability amplitudes.


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