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Taking 2 time-dependant states $\vert s_1 \rangle , \vert s_2 \rangle$ , you can do the following calculation \begin{align*} i \hbar \partial_t \langle s_1 \vert s_2 \rangle &= \big\{i \hbar \partial_t \langle s_1 \vert \big\} \vert s_2 \rangle + \langle s_1 \vert \big\{ i \hbar \partial_t \vert s_2 \rangle \big\} \\ &= -\langle s_1 \vert H \vert s_2 \rangle + \langle s_1 \vert H \vert s_2 \rangle \\ &= 0\end{align*} Where $ i \hbar \partial_t \vert s \rangle = H \vert s \rangle $ and $ -i \hbar \partial_t \langle s \vert = \langle s \vert H $ were used in the process.

If the two states are identical (let's say $\vert s \rangle$), this implies that the state stays normalized over time, meaning the probability of finding the system in any state (that $\vert s \rangle$ has in its linear combination) stays 1 .

But from the above you can follow more generally $$ \vert \langle s_1 \vert s_2 \rangle \vert^2 ~~\text{is constant over time} $$ Which I would not have expected, based on some intuition a priori. And I still don't have any after it. So my question is: Is there a physical interpretation of this result? (like in the case of two identical states) And if yes; Is there an intuitive reason why it should be true?

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  • $\begingroup$ Simple application of polarization identity... $\endgroup$ Dec 21 '16 at 19:29
  • $\begingroup$ Well, the parts of $s_1$ and $s_2$ which are outside of their overlap $s$ do not really survive the scalar product. So the scalar product is pretty much like the one you considered for identical states, when it comes to the action of the Hamiltonian on it, no? $\endgroup$ Dec 22 '16 at 1:42
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Time evolution is a linear map $$x \to U_t x = x(t)\:.$$ Above $U_t = e^{-it H/\hbar}$ giving rise to the standard Schroedinger equation. Since scalar products in complex vectors spaces satisfy $$4\langle x| y\rangle = ||x+y||^2 - ||x-y||^2 - i ||x+iy||^2 +i ||x-iy||^2\:, $$ if the linear map $U_t$ preserves the norms (as a consequence of preservation of total probability), then it also preserves the scalar products.

Due to the observation above, making use of linearity of time evolution referring to the superposition principle of pure states, i.e., $u(t)+v(t) = (u+v)(t)$, we have $$4\langle x(t)| y(t) \rangle = ||x(t)+y(t)||^2 - ||x(t)-y(t)||^2 - i ||x(t)+iy(t)||^2 +i ||x(t)-iy(t)||^2$$ $$= ||(x+y)(t)||^2 - ||(x-y)(t)||^2 - i ||(x+iy)(t)||^2 +i ||(x-iy)(t)||^2$$ $$= ||x+y||^2 - ||x-y||^2 - i ||x+iy||^2 +i ||x-iy||^2 =4\langle x| y \rangle$$

The fundamental reason is of geometric nature: As soon as we introduce the scalar product in QM and assume that the relevant norm in QM, that related with probabilities, is the one associated with that scalar product, we can exploit the well-known geometric fact every notion of scalar product can be construct out of the associated notion of norm.

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  • $\begingroup$ Also, $\langle s_1 | s_2 \rangle \to \langle s_1 | U_t^\dagger U_t | s_2 \rangle = \langle s_1 | s_2 \rangle$ if we assume the knowledge that $U_t$ is unitary. (Of course, you just proved that every norm preserving map is unitary.) $\endgroup$
    – Noiralef
    Dec 21 '16 at 23:18
  • $\begingroup$ Actually it is false that norm preserving implies unitary. It just implies scalar product preserving (the proof is the one I wrote above) which is less than unitarity as unitarity also includes surjectivity. $\endgroup$ Dec 22 '16 at 6:42
  • $\begingroup$ This shows that invariance of the norm implies invariance of the scalar product. But this does not answer the question. The question being : Why should we expect $\vert \langle s_1 \vert s_2 \rangle \vert^2$ to be constant over time? For $\vert \langle s \vert s \rangle \vert^2$ there is an intuition on why it should be constant, but is there also an intuitive way to understand why $\vert \langle s_1 \vert s_2 \rangle \vert^2$ is constant? $\endgroup$
    – user122721
    Dec 22 '16 at 10:25
  • $\begingroup$ Because one can construct the scalar product by means of sum an difference of lengths (norms) of vectors, exactly as it happens in elementary geometry where $4 x \cdot y = (x+y)\cdot (x+y)- (x-y)\cdot (x-y)$ and time evolution is linear. This is the only possible intuition in my view. $\endgroup$ Dec 22 '16 at 10:45
  • $\begingroup$ I made more explicit my answer... $\endgroup$ Dec 22 '16 at 10:51
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First of all, it's not just the case that the norm-squared $| \langle s_1 | s_2 \rangle|^2$ is constant in time; in fact the inner product $\langle s_1 | s_2 \rangle$ itself is constant, which is a much stronger statement.

One way to understand this is to think in the Heisenberg picture, where the states do not evolve in time but the operators do. An inner product $\langle s_1 | s_2 \rangle$ is just a matrix element of the identity operator $\langle s_1 | \hat{I} | s_2 \rangle$. The fact that inner products do not evolve over time simply reflects the fact that the identity operator $\hat{I}$ is time-independent for any Hamiltonian. This is somewhat intuitive - you would expect that the identity operator should always have value 1 in any possible situation.

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