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I always have trouble remembering the sign in factors like $\exp(\pm ik\cdot x)$ (I'll use mostly minus signature here) that arise in field theory.

My mnemonic is to remember that the Schrodinger equation is $i\partial_t|\Psi\rangle=H|\Psi\rangle$ and so the wavefunction goes something like $|\Psi\rangle\sim \exp(-i\omega t)$ where $\omega$ is the energy of the state and then I just remember that this is the same sign assigned to incoming particles in Feynman diagrams. Similarly, this is the same sign that comes with the annihilation operator when we expand out a field in terms of creation and annihilation operators. The sign in $|\Psi\rangle\sim \exp(-i\omega t)$ is then just determined by the sign appearing in the Schrodinger equation above.

My question is whether the sign appearing in the Schrodinger equation is physical? Relatedly, are the signs appearing in Feynman diagrams or more generally the overall sign in the path integral physical?

It would seem to me that all calculations would remain the same if we switched signs everywhere since physical quantities just depend on complex inner products and this would appear to be insensitive to flipping the signs everywhere..

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  • $\begingroup$ Flipping the sign is the same thing as reversing the direction of time. You can detect this microscopically if you don't also flip charges and parity. You can always detect this macroscopically via (say) the second law of thermodynamics. $\endgroup$ – Ryan Thorngren Dec 19 '13 at 6:32
  • $\begingroup$ What led to Schrodinger to choose the sign in his equation, though? How would he know the sign above corresponds to the proper time direction? $\endgroup$ – user26866 Dec 19 '13 at 7:39
  • $\begingroup$ Related: physics.stackexchange.com/q/9557/2451 and links therein. $\endgroup$ – Qmechanic Dec 19 '13 at 9:46
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There is no physical significance because there are two equally good square roots of negative one, i.e. $\pm i$ and unlike $\pm\sqrt 2$ there is no way to tell the roots apart.

So what you call $+i$ and what I call $-i$ could actually be the same. What matters is relative signs, for instance $e^{i(kx-\omega t)}$ is a wave traveling to the right.

You could also write $e^{j(\omega t-kx)}$ to be a wave traveling to the right (with $j^2=-1$).

The first gives $-i\omega$ for a time derivative and the latter gives $j\omega$ for a time derivative but they could be the same thing if $j=-i$ so if you like to write one or like to have time derivatives do one then that could tell you which to favor.

It's like the question about whether my subjective experience of red is the same as your red subjective experience of red.

If you like to write your time terms like $e^{-i\omega}$ and I like to write them like $e^{j\omega}$ they could be same thing since there is no way to tell the difference between two square roots of $-1$ there are two and they are negatives of each other, but what I call i might be what you call negative $i.$ But we can write i and write $-i$ and tell them as negatives of each other and that is all that matters.

So be consistent so we know when you are writing about different roots or when you are writing about just one.

So I'm saying is that you are right, but you can just flip them in your mind. Multiplication by $ i$ could rotate by 90 degrees clockwise and you could write the positive $i$ axis as the negative y axis and nothing would change in the math but you would technically mean by $i $ what your friend means by $-i$.

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You are correct, but one place where it tends to bite you is in Fourier transforms. With Fourier transforms you really want the inverse transform to have $+i$ in it, so that $\delta(k - k_0)$ becomes $e^{i\,k_0\,x}$ when you transform back to position space. (It doesn't have to be, but that's the easiest to remember: the delta at $k_0$ produces $e^{i\,k_0\,x}$.)

Once you've chosen this, your momentum operator needs to be $p = \hbar ~k$; operating on the wavefunction $f(x) = \int dk~e^{i\,k\,x}~f[k]$ this should by all rights be $-i\, \hbar~\partial_x$ to match. In addition, a wave with positive momentum should move forwards, giving $\exp(i\,k\,x - i\,\omega\,t)$, so deriving the energy $E = \hbar~\omega$ should be done with $i\,\hbar ~\partial_t$.

You can do it the other way as long as you start thinking about $e^{i k_0 x}$ as pointing "backwards" rather than "forwards".

In electrical engineering texts it is very common to see the reverse convention as well as the variant letter $j$; (because $i$ refers to currents too often). For example, Staelin, Morgenthaler, & Kong's Electromagnetic Waves is very fond of $\exp(j\,\omega\,t)$ terms. If you have to juggle conventions between these two disciplines it starts to become very helpful to mentally define $j = -i$ to cross-interpret.

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