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

If we 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 $\boldsymbol{x}$?

(Added) (I had drafted but missed while copy pasting)

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". I wonder why this could not be possible.

Clarification (after answer by Alfred Centauri)

My question is why consider the evolution at all (What ever the case may be and what ever the parameter may be, time or proper time or whatever).

My motivation here is to study the nature of the theory of quantum mechanics as a statistical model. I am looking at it from that angle.

  • $\begingroup$ Related: physics.stackexchange.com/q/32166/8563 $\endgroup$ – Emilio Pisanty Jul 19 '12 at 12:47
  • $\begingroup$ So do I understand you to be asking for a block world formulation of quantum theory? For which you could use the Wightman axioms (albeit they're not close to the successes of Lagrangian QFT). They introduce a single Hilbert space that supports a representation of the Poincaré group, and time is not privileged over space (except for the 1+3 signature). Lagrangian QFT somewhat obscures a block world perspective, insofar as it focuses on a Hilbert space at a single time, corresponding to phase space observables, however a block world perspective of Lagrangian QFT is possible. $\endgroup$ – Peter Morgan Jul 19 '12 at 20:25
  • $\begingroup$ @RajeshD: The Heisenberg formulation takes your point of view, the wavefunction is time independent, but the observables depend on time. This just means that the interaction with the particle at different times is by different operators. $\endgroup$ – Ron Maimon Jul 20 '12 at 5:24

I think the main reason is practical, but it might be related to a theoretical reason.

The main reason is that we almost never use the time-dependent Schroedinger equation because if the state wasn't stationary, its rate of change would be, at the usual atomic scales, so fast that we couldn't measure it or study it empirically with laboratory-sized apparatus. Similarly, what governs the observable properties of macroscopic bodies, such as their chemical bonds and colours, involves stationary states. If the states weren't stationary, the body would not persist long enough for us to consider it as having a property. It is striking how little direct empirical support the time-dependent Schroedinger equation has, and how little use it finds. We don't even use it to study scattering events (which, admittedly, for a very brief time occur very rapidly).

This might be related to a deeper theoretical reason one finds in statistical mechanics. In statistical mechanics, it is often pointed out that measurements made with laboratory-sized equipment necessarily involve a practically infinite time average such as $$\lim_{T\rightarrow\infty}\frac1T\int_0^T f(t)g(t)dt.$$ Well, in Quantum Mechanics, measurement has something similar about it, in that it always involves amplification of something microscopic up to the macroscopic scale so we can observe it (an observation made by many, including Feynman), and the main way to do this seems to be to let the microscopic event trigger the change from a meta-stable state to a stable equilibrium state of the laboratory-sized apparatus (H.S. Green Observation in Quantum Mechanics, Nuovo Cimento vol. 9, pp. 880--889, posted at http://www.chicuadro.es , and many others since). Once again, this involves a long-time, stable, equilibrium as in Statistical Mechanics. But the relation to the practical reason is not completely clear.

That said, in theory it is sometimes possible to rephrase the time-dependent Schroedinger evolution equation into a space-evolution equation, even though no one ever does this since it has no earthly use. Consider the Klein--Gordon equation (which is the relativistic version of Schroedinger's equation), $$({\partial\over \partial x}^2-{\partial\over \partial t}^2 + V )\psi = 0.$$ Obviously, we can get isolate either $x$ or $t$, and under certain conditions take the square root of the operator to get $$ {\partial\over \partial x} \psi = \sqrt{ ({\partial\over \partial t}^2 - V)}\psi .$$

Under the usual physical assumptions of flat space--time and no field-theoretic effects, one could do this to isolate $t$ and get the time evolution because we assume that energy is always positive, so we can indeed take the square root (all the eigenvalues of the Hamiltonian are positive). This may not always be true when, as here, we try to isolate $x$ and get the space-evolution.


Now, as to the question of why consider any evolution at all, why not just consider $\psi(x,y,z,t)$ in a relativistically timeless fashion, the main answer is that it wreaks havoc with the idea of measurement, observable. and the justification of the Born interpretation. Dirac tried to write a Quantum Mechanics textbook your way, but gave up even after the fifth chapter, where he remarks that the notion of observable in not relativistic, and for the rest of the book he proceeds non-relativistically (until he get to the Dirac Equation at the end). The second edition abandons the attempt to be relativistic, is more traditional and uses the time evolution point of view from the start. He remarked, famously,

The main change has been brought about by the use of the word «state» in a three-dimensional non-relativistic sense. It would seem at first sight a pity to build up the theory largely on the basis of nonrelativistic concepts. The use of the non-relativistic meaning of «state», however, contributes so essentially to the possibilities of clear exposition as to lead one to suspect that the fundamental ideas of the present quantum mechanics are in need of serious alteration at just tbis point, and that an improved theory would agree more closely ' with the development here given than with a development which aims at preserving the relativistic meaning of «state» throughout.

And in fact Relativistic Quantum Mechanics, as opposed to field theory, is, like many-particle Relativistic (classical) mechanics, not theoretically very well developed. There seem to be so many problems, people prefer to jump right to Quantum Field Theory in spite of the divergences and need for renormalisation and everything. Furthermore, relativistic QM is restricted to the low energy regime since with high energies, particle pair production is possible, yet the equations of QM hold the number of particles as fixed and do not allow for pair production.

  • $\begingroup$ Thanks for the nice answer. It was a joy reading it. You really got the spirit of the question. $\endgroup$ – Rajesh Dachiraju Jun 7 '13 at 22:39

(1) In the Heisenberg picture, the wavefunction does not evolve with time, the operators do.

(2) For relativistic covariance, $t$ ought to be a coordinate with proper time $\tau$ as the evolution parameter.

(3) In QFT, which is relativistically co-variant, $t$ is a coordinate.

If these don't begin to address your question, please re-edit your question to clarify.

  • $\begingroup$ I have edited with a clarification in view of your answer. $\endgroup$ – Rajesh Dachiraju Jul 19 '12 at 13:10

it's an empiral fact that time exists, and states evolve in time. or is that really the case, or does it just seem so? interesting question. anyway, feynman path integrals, no such problem.

  • $\begingroup$ Sorry I missed a crucial part of the question while copy pasting the draft. Now I have added it. I hope you excuse this. $\endgroup$ – Rajesh Dachiraju Jul 19 '12 at 12:39

You can, sort of. You can take $\psi(x)$ to satisfy the time-independent Schrödinger equation, for some eigenvalue $E_n$ of the Hamiltonian operator that appears in the time-dependent Schrödinger equation. However I would take that to make the time-independent formalism less fundamental. It's also possible for the time-dependent state to be in a superposition of different energy states, which doesn't play well with the time-independent formalism.

  • 1
    $\begingroup$ I think you have digressed a bit from what I had in mind. I do not suggest to consider time independent Schrodinger equation. I am not interested in that and that is not the only choice. My question is just why consider evolution of wave function at all? $\endgroup$ – Rajesh Dachiraju Jul 19 '12 at 13:00

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