How do you derive the Schrödinger equation (wave mechanics, time dependent state) from Heisenberg's Matrix Mechanics (matrix based, time dependent operators)


The basic idea is simple enough: one seeks a unitary transformation on the Hilbert space that preserves all physical measurements, which is equivalent to keeping entities of the form $\langle \psi,\,\hat{A}\,\psi\rangle$ unchanged, where $\psi$ is a quantum state and $\hat{A}$ an observable (or any of its powers). Furthermore, we want to "freeze" the states. So suppose we begin with a Schrödinger picture state $\psi$ evolving by:

$$i\,\hbar\,\partial_t\,\psi = \hat{H}\,\psi$$

or, equivalently, $\psi(t) = \exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0)$. We can easily justify the the existence of some self-adjoint $\hat{H}$ from any unitary one parameter group of evolutions in the finite dimensional case; in the infinite dimensional case, one appeals to the Stone Theorem on One Parameter Unitary Groups.

So now, how do our physical measurements vary? They vary like:

$$M(t) = \langle \psi(t),\,\hat{A}\,\psi(t)\rangle$$

when we write them down in the Schrödinger picture. But now:

$$\begin{array}{lcl}M(t) &=& \langle \psi(t),\,\hat{A}\,\psi(t)\rangle\\ &=& \langle \exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0),\,\hat{A}\,\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0)\rangle\\ &=&\langle \psi(0),\,\left(\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\right)^\ast\,\hat{A}\,\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0)\rangle\\ &=&\langle \psi(0),\,\exp\left(+\frac{i}{\hbar}\,\hat{H}\,t\right)\,\hat{A}\,\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)\,\psi(0)\rangle \end{array} $$

so that we'll keep all our measurements unchanged with frozen states if and only if our observables evolve following:

$$\hat{A}(t) = \exp\left(+\frac{i}{\hbar}\,\hat{H}\,t\right)\,\hat{A}\,\exp\left(-\frac{i}{\hbar}\,\hat{H}\,t\right)$$

so this is the unitary transformation of observables we seek. Let's work out the differential form of this one: we get:

$$\partial_t\hat{A}(t) = \frac{i}{\hbar}\,\left(\hat{H}\,\hat{A}(t)-\hat{A}(t)\,\hat{H}\right) = \frac{i}{\hbar}[\hat{H},\,\hat{A}]$$

which of course is the Heisenberg equation, the quantum analogue of the Liouville equation from Hamiltonian mechanics.

Going the other way is straightforward; one uses the inverse transformation . Or one can derive it from scratch using the same principle as above: we must keep our measurements fixed.

  • $\begingroup$ But what if I wanted to say, derive wave mechanics completely from Heisenberg's matrices, including the form of the operators? $\endgroup$ – user140323 Mar 20 '18 at 13:06
  • $\begingroup$ Perhaps you could use the Stone-Von Neumann theorem, which says that up to unitary equivalence, there is only one choice for the position and momentum operators that satisfy the commutation relation. $\endgroup$ – user1379857 Mar 20 '18 at 13:34
  • $\begingroup$ @user1379857 Yes that would certainly do it if the OP wanted to take the CCR as fundamental. But I'm not really sure what he/she wants to begin from. $\endgroup$ – WetSavannaAnimal Mar 20 '18 at 13:47
  • $\begingroup$ @user140323 See my comment above on Stone-von Neumann: I'm unsure exactly what you want to begin from and perhaps you should rewrite your question to state it. $\endgroup$ – WetSavannaAnimal Mar 20 '18 at 13:48

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.