Let's have wave-function $\lvert \psi \rangle$. The full probability is equal to one:

$$\langle \Psi\lvert\Psi \rangle = 1.\tag{1}$$

We need to introduce time evolution of $\Psi $; we know it in the initial moment of time. So it's naturally set

$$\lvert \Psi (t) \rangle = \hat {U}|\Psi (0) \rangle ,$$

and from $(1)$ it follows that

$$\hat {U}^{\dagger}\hat {U} = \hat {\mathbf E}.$$

So it may be represented as $U = e^{i \alpha \hat {H}t}$ (we suppose that $\hat {H}^{\dagger} = \hat {H}$ and $\hat {H} \neq \hat {H}(t)$ for simplifying derivation). Thus it is possible to write

$$\partial_{t}\lvert\Psi (t) \rangle = i\alpha \hat {H}| \Psi\rangle.$$

But how to get the physical interpretation for $\hat {H}$?

  • 1
    $\begingroup$ I believe this is part of how Dirac showed the equivalence of Heisenberg's matrix mechanics and Schrodinger's wave mechanics, and I suspect your missing ingredient is the canonical commutator relation. $\endgroup$ – David H Nov 4 '13 at 22:54
  • $\begingroup$ Why is this in the close vote queue for unclear questions?! The question is very clear; how one gets the physical interpretation for the $\hat{H}$. $\endgroup$ – Abhimanyu Pallavi Sudhir Nov 5 '13 at 2:38
  • $\begingroup$ @DIMension10: It's always possible that flagged posts may peek into the queue sometimes, unless it's very serious :) $\endgroup$ – Waffle's Crazy Peanut Nov 5 '13 at 9:15
  • $\begingroup$ @DavidH . This may be showed by using postulate of using Hamilton's formalism in quantum mechanics. But method I described doesn't use it. At least explicitly. $\endgroup$ – John Taylor Nov 5 '13 at 14:29

One can indeed motivate the Schrödinger equation along the lines you suggest. The crucial point you are missing is time shift invariance of your quantum system. It is this that lets you write down $U = \exp(i\alpha\,H\,t)$.

To explain further:

  1. The fact of $\psi(t) = U(t) \psi(0)$ is simply a linearity assumption.

  2. The evolution wrought by your state transition matrix $U(t)$ over time n units is simply the matrix product of the individual transition operations over 1 unit, so $U(n) = U(1)^n$ for integer $n$. This is simply saying the evolution operator for any fixed time interval is the same, no matter when it is imparted to the quantum state (i.e time shift invariance). The evolution in the same experiment doesn't care whether I do it now or in ten minutes time after I have my cup of tea (as long as I don't spill any tea on the experiment, say). It's a Copernican notion. Arguing like this, and slicing the time interval $t$ in different ways, you can quickly prove things like $U(p) = U(1)^p$ for all rational $p$ and $U(t+s) = U(s)U(t) = U(t) U(s);\,\forall s,t\in\mathbb{R}$. The only continuous matrix function with all these properties is $U = \exp(K t)$, for some constant $K$.

  3. Now the assumption of probability conservation ("full possibilities") is brought to bear. This means that $U$ must be unitary - your $U^\dagger U = U U^\dagger = I$. This means $\exp(K t) \exp(K^\dagger t) = \exp(K^\dagger t) \exp(K t) = I$. So $K$ and $K^\dagger$ must commute and $K + K^\dagger = 0$, whence $K = -K^\dagger$. $K$ is thus skew Hermitian. Now every skew Hermitian $K$ can be written as $K = i H$, for some Hermitian $H$. And we can pull out any nonzero, real scalar factor we like to get $U(t) = \exp(i\alpha H)$

The rest of your argument follows. How do we get the physical interpretation for $H$? Its simply a hunch. With a bit of work (converting from the Schrödinger to Heisenberg picture) you can show that any observable that commutes with $H$ is constant in time - its eigenvalues do not evolve. $H$'s eigenvalues are thus. So, if we postulate that $H$ is indeed an observable $\hat{H}$ with its full blown recipe for measurement interpretation instead of simply a plain boring old operator, then all the statistics of its measurements are constant in time. It represents a time-shift-invariant symmetry. So, by analogy with the classical energy and Noether's theorem for the classical energy, we simply postulate that $H = \hat{H} = {\rm energy\, observable}$.

Richard Feynman uses exactly this approach in the early chapters of Volume 3 of his Lectures on Physics. There is a chapter 7 called "The Dependence of Amplitude on Time" wherein he does this much better than I do.

  • $\begingroup$ arguably the best answer I've read. $\endgroup$ – Kyle Kanos Nov 5 '13 at 0:47
  • $\begingroup$ Thank you for very impressive answer! It helped very much. $\endgroup$ – John Taylor Nov 5 '13 at 4:46
  • 1
    $\begingroup$ @JohnTaylor It was a pleasure. Richard Feynman deserves most of the credit here - check out his third volume if you're interested in stuff like this. He's a bit allergic to the Heisenberg picture for some reason, which he describes as the quantum mechanics that some people do if they feel like doing problems in rotating frames of reference and neglects it in his lectures (although he kind of takes it up in the chapter "Symmetry in Physics"). Not that he had any problems working with this picture when needed. $\endgroup$ – WetSavannaAnimal Nov 8 '13 at 8:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.