# Multiplying wavefunction with complexnumber.

• I
Gold Member
Is it true that by multiplying wavefunction with arbitrary complexnumber, which module is 1, results another wavefunction, that has same physical meaning? aka ##\forall_\phi(\Psi\ has\ the\ same\ meaning\ as\ \Psi \cdot e^{i \cdot \phi})##
If not please give me an example of wavefunction and ##\phi## where it does not.

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Metmann
##\Psi## and ##e^{i\phi}\Psi## represent the same state since ##|\Psi|^2## is the central quantity that characterizes the state.

Gold Member
##\Psi## and ##e^{i\phi}\Psi## represent the same state since ##|\Psi|^2## is the central quantity that characterizes the state.
Yes, but does ##e^{i\phi}\Psi## satisfy Schrödinger equation with same potentialenergy function ##U(t;\vec X)## as ##\Psi##?

Metmann
Yes, but does ##e^{i\phi}\Psi## satisfy Schrödinger equation with same potentialenergy function ##U(t;\vec X)## as ##\Psi##?
Sure. The Schrödinger equation is ##\mathbb{C}##-linear on both sides, hence the phase drops out. In fact all quantum mechanics is ##\mathbb{C}##-linear, hence the physical space of states is the projective Hilbert space.

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vanhees71 and olgerm
bhobba
Mentor
Yea - all true.

Why - because states are not really members of a Hilbert space - they are really positive operators of unit trace. Such operators of the form |u><u| are called pure and can be mapped to the underlying space the operator is defined on by simply using the u - but note if you multiply |u> by a complex number of unit length to get |u'> |u><u| = |u'><u'| - that's why it is invariant to phase. In general any positive operator of unit trace can be put in the form (not necessarily uniquely BTW which has implications for the decoherence program in explaining the measurement problem I will not go into here - start another thread if interested) as U=∑pi |ui><ui| where pi are positive and sum to 1. These are called mixed states. The Born Rule then becomes the expected value of an observable O, E(O), is E(O) = trace (OU) where U is the systems state and show the pi in fact are the probability of Iui><ui| in the mixture.

In fact due to a very famous theorem by the mathematician Gleason it can be deduced from more fundamental assumptions - see post 137:

Thanks
Bill

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vanhees71
DrClaude
Mentor
Yes, but does ##e^{i\phi}\Psi## satisfy Schrödinger equation with same potentialenergy function ##U(t;\vec X)## as ##\Psi##?
Why don't you try it yourself? Write down the Schrödinger equation for ##\Psi##, make the substitution ##\Psi \rightarrow ^{i\phi}\Psi##, and see if anything changes.