Since the time evolution unitary operator preserves norm, if applied to any system say electron whizzing around its orbitals, no matter what time we consider would it always have the same probability amplitude and distribution?


In Quantum Mechanics, within the Copenhagen interpretation, the squared modulus of the wavefunction is associated to the probability of finding a particle in a certain position in space. Since, according to an axiomatic definition of probability, a correctly defined probability is normalized, in this case we can say that also the squared modulus of the wavefunction is normalized. This has two consequences:

  • The integral of the squared modulus of the wavefunction on a sub-region of the whole space is lower than or equal to one;

  • The integral of the squared modulus of the wavefunction on the whole space is identically equal to one.

We can now consider how this probability distribution, associated to the squared modulus of the wavefunction, is evolving in time. As you correctly stated, the time evolution operator $$ U(t) = e^{-i\frac{\mathcal{H}}{\hbar} t} $$ where $i$ is the imaginary unit, $\mathcal{H}$ is the hamiltonian of the system, $\hbar$ is the reduced Planck constant and $t$ the time, is a unitary operator. This means that it will preserve the norm of the wavefunction that is, actually, the squared modulus of the whole wavefunction. This has an important consequence: for any subsequent time instant, the integral of the squared modulus of the wavefunction over the whole space is one. From a physical point of view, this means that we are again sure to find the particle in the considered space also after the application of the time evolution operator, allowing to make use, again, of the same interpretation, giving a physical meaning to the quantity we obtain. It is important to notice that this does not mean that the wavefunction itself (and the associated distribution) has not changed during time. After a certain time interval $t$, the wavefunction associated can be different, the wavepacket associated to a given particle can have spread and its center can have moved from point $A$ to point $B$, but we know that integrating the squared modulus of the wavefunction over a certain region of space we can obtain again the probability of finding the particle in that region.

To conclude, my answer only deals, in terms of observables, with the position of the particle, but the same reasoning can be applied to other observables and to the associated operators. Moreover, I only adopted a Schrödinger picture, where the time evolution operator is applied to the wavefunction determining its evolution, but the same can be done in the Heisenberg picture, where the time dependence is instead incorporated in the observables.

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  • $\begingroup$ All well and good, but scattering states are usually not normalizable. Also CW Laser states ar not described by normalizable states. So some ppl say the right approach is a rigged hilbert space (with more bras than kets) $\endgroup$ – lalala Oct 11 '19 at 9:40

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