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I was hoping that someone could give me the more fundamental reason that we take as the temporal part of a quantum wavefunction the function $e^{-i\omega t}$ and not $e^{+i\omega t}$? Clearly $e^{-i\omega t}$ solves the time dependent Schrödinger equation whereas $e^{+i\omega t}$ does not.

However, the Schrödinger equation, when it was first developed, was merely a hypothesis. It was new physics and, as such, could not be derived from previous work. Hence, why did Schrödinger and his contemporaries choose $e^{-i\omega t}$ and, thus, why does an antiparticle with wavefunction temporal dependence $e^{+i\omega t}$ correspond to backwards time travel or negative energy?

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up vote 6 down vote accepted

In mathematics, there is a complete symmetry between $+i$ and $-i$. Both the imaginary unit and the minus imaginary unit obey $$ i^2 = (-i)^2 = -1 $$ The exchange of $i$ and $-i$ is known as the ${\mathbb Z}_2$ automorphism group of the complex numbers ${\mathbb C}$. When you introduce the complex numbers for the first time, it's a complete convention whether you call a square root of $(-1)$ as $+i$ or $-i$.

However, in physics, we have to break the symmetry between $+i$ and $-i$ because we must know whether a wave in a particular situation is $\exp(i\omega t)$ or $\exp(-i\omega t)$, for example. In particular, $xp-px=i\hbar$ and not $-i\hbar$. Also, and the following choice of the sign is actually not independent from the previous one in the commutator, Schrödinger's equation was chosen to be $$ H |\psi \rangle = i\hbar\frac{d}{dt}|\psi\rangle $$ where $H$ is the Hamiltonian that may be replaced by $H=E$ when acting on an energy eigenstate $|\psi\rangle$. This equation is totally universal everywhere in quantum mechanics where a Hamiltonian is well-defined (it may be even quantum field theory or some descriptions of string theory).

The equation above, with $H=E$, is solved by $$|\psi(t)\rangle = \exp(Et/i\hbar) |\psi(0)\rangle = \exp(-iEt/ \hbar) |\psi(0)\rangle = \exp(-i\omega t) |\psi(0)\rangle $$ All the forms are equivalent because $1/i = -i$ – this equation is equivalent to $i^2=-1$ – and because $E=\hbar\omega$ without a minus sign. So your sign is wrong; the sign you denounced is the right one and the sign you wanted is the incorrect one.

Just to be sure, in quantum field theory, we work with various objects – quantum fields – that are expanded into terms that depend on time as $\exp(-i\omega t)$ while there must also be terms that depend on time via $\exp(+i\omega t)$. But these are terms in operators, not the time dependence of the wave function. One must be careful about the precise statements and objects. I haven't made any statement of the sort that only the expression $\exp(-i\omega t)$ and not $\exp(+i\omega t)$ appears in quantum theory papers and books. Of course, both of them may appear somewhere – in quantum field theory, both of them have to appear because there are both creation and annihilation operators, both particles and antiparticles. But when we are asking how an energy $E$ wave function (and I mean the ket vector) depends on time, it's always via $\exp(-iEt/\hbar)$. The bra vector has the opposite sign (plus) in the exponent.

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And I was thinking it was simply a normalization and convergence issue... – Dylan Sabulsky Dec 7 '12 at 5:32

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