Why do we assume that the wave function should satisfy the Schrödinger equation? If a function satisfies the Schrödinger equation, does it mean that it is a wave function?
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$\begingroup$ I don't believe we assume that the wave function satisfies the schrodinger equation... We assume a form of solution to schrodinger's equation based on the resultant differential equation of the Hamiltonian; the total energy of the system. The solutions of psi are derived based on the Hamiltonian and time independent schrodinger equations. The Hamiltonian acts on psi, revealing the energy eigenvalues of psi. Wavefunction psi is found from solving the differential equation of the time independent schrodinger equation. $\endgroup$– bleuofblueCommented Dec 22, 2016 at 1:43
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3 Answers
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Because wave functions satisfying the Schroedinger equation adequately describe experimental data.
I am not sure a function satisfying the Schroedinger equation is necessarily a wave function of quantum theory. For example, such a function can be used to describe Couder's experiments with bouncing droplets (https://arxiv.org/abs/1401.4356)
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$\begingroup$ If a function satisfies the Schrödinger equation, does it mean that it is a wave function? $\endgroup$ Commented Dec 21, 2016 at 13:59
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$\begingroup$ It means it is a wave function, but the next question is 'Of what Hamiltonian?'. $\endgroup$ Commented Dec 21, 2016 at 15:49
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$\begingroup$ A function satisfying the Schrodinger equation is a wave function only if it can be normalized. $\endgroup$ Commented Dec 21, 2016 at 16:04
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$\begingroup$ @ZhenyuYuan : IMHO, not necessarily. One reason is given in my answer. $\endgroup$ Commented Dec 21, 2016 at 19:00
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If a function satisfies the Schrödinger equation, does it mean that it is a wave function?
If, by wavefunction, you mean a position basis representation of a quantum state, then the answer is no.
For example, there are a continuum of solutions to the quantum harmonic oscillator time independent Schrödinger equation
$$\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{m\omega^2}{2}x^2\right)\psi(x)=E\psi(x)$$
but only a countably infinite subset of these solutions are normalizable and thus are wavefunction representations of states. See, for example, this Wolfram Demonstration
answered Dec 21, 2016 at 23:34
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Because Schrödinger's equation is derived from a solution of the classical wave equation. I'm assuming this is what you mean by the wave function i.e. the vacuum solutions. That is how Schrödinger himself derived the equation. Check this paper out.
Basically, you start with the wave solutions in vacuum, substitute in the classical wave equation, and use the energy and momentum of a photon. The end result is Schrödinger's equation.
Note how the energy operator is derived from the time components, and the momentum operator from the spatial components. This fits nicely with Noether's theorem.
