# Why do we consider the evolution (usually in time) of a wave function ?

#### rajesh_d

Why do we consider evolution of a wave function and why is the evolution parameter taken as time, in QM.

Look at a simple wave function $\psi(x,t) = e^{kx - \omega t}$. $x$ is a point in configuration space and $t$ is the evolution parameter. They both look the same in the equation, then why consider one as an evolution parameter and other as configuration of the system.

My question is why should we even consider the evolution of the wave function in some parameter (it is usually time)?. Why can't we just deal with $\psi(\boldsymbol{x})$, where $\boldsymbol{x}$ is the configuration of the system and that $|\psi(\boldsymbol{x})|^2$ gives the probability of finding the system in the configuration $\boldsymbol{x}$?

One may say, "How to deal with systems that vary with time?", and the answer could be, "consider time also as a part of the configuration space". Why wonder why this could not be possible.

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#### Simon Bridge

Homework Helper
Why do we consider evolution of a wave function and why is the evolution parameter taken as time, in QM.
Because we want to know what happens next. Generally a system will change in time... so the square modulus of the wavefunction only give the correct probability-density at a particular point in time.

Yes we can work time into the configuration space - this is the point of relativistic quantum mechanics.

Even so, we have to do experiments in a laboratory which is mired in time - so the coordinate that is usually most convenient to evolve the wavefunction with respect to is still time.

#### vanhees71

Gold Member
Time in quantum theory is necessarily a parameter labelling the "causal sequence of events" or something like that, no matter whether you consider non-relativistic or relativistic quantum theory.

The reason has been given by Pauli very early in the development of quantum theory: If time would be taken as an observable than it would have the commutation relation
$$[\hat{t},\hat{H}]=\mathrm{i}\hbar$$.
This would mean, as for the case of the position and momentum operator in non-relativistic quantum theory, that the spectrum of both the time and the Hamilton operator would be whole $\mathbb{R}$, and this contradicts the stability of matter since for that the energy must have a lower boundary, so that a ground state of lowest energy exists.

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