Does swapping "$i$" with "$-i$" change a physical theory?

Question

Mathematically speaking, "$i$" and "$-i$" are the two roots of the equation $x^2+1=0$ and it seems to me at least that there is no obvious way of distinguishing between them. Thus, what we call "$i$" and "$-i$" is a matter of convention. If this is the case, I wonder if one can replace all the "$i$"s in a physical theory by "$-i$"s and get back all the same physics provided the change has been made consistently?

To be concrete, the canonical example is the Schrödinger equation: if I make the above mentioned change, the TDSE reads $$-i\hbar \frac{d}{dt}|\psi\rangle = H |\psi\rangle$$ which would be equivalent to reversing the arrow of time. I don't know too much about this but I have heard of phase transitions that break time-reversal symmetry. In that case, the physics indeed changes if I swap "$i$" with "$-i$". How is that possible?

Answer 1

Changing $i$ to $-i$ is not the same as changing the arrow of time.

Changing $i$ to $-i$ is entirely irrelevant, it has no effect on the physics. It would be like inventing a new symbol for the number 3, and using that symbol instead of "3". The symbol itself is just a convention. Nothing really depends on it. The formulas look slightly different if you use different conventions. But physics remain the same. But, as stressed in the OP, using a different convention is allowed only if you are consistent. You would not, for example, use the symbol "4" to denote the number 3, and vice-versa, but make this change only in half your formulas. If you redefine $i\leftrightarrow-i$, you better change it everywhere.

Changing the arrow of time is not irrelevant at all. Some theories are invariant under time-reversal, and changing the arrow for those is fine. It would be like using a rotated frame of reference for a system that is invariant under rotations. Equations would look the same.

But not all theories are invariant under time-reversal. If you send $t\to-t$ in those, you would be changing the physics of the problem. So it is clear that the redefinition $i\leftrightarrow -i$ (the Galois automorphism of the extension $\mathbb C/\mathbb R$) is not the same as reversing time. The former is a matter of conventions. The latter is a physical operation, which may or may not leave the physics invariant. Sometimes it does, but most often it doesn't.

What is reversing the arrow of time anyway?

Imagine you want to figure out whether a given system is invariant under rotations. How do you do this? You can rotate the system, while keeping your measuring apparatus fixed. If you get a different answer, the system is not invariant under rotations. Of course, you could equivalently rotate the apparatus and keep the system fixed, for the same result. But, clearly, you wouldn't rotate both. If you did that, you wouldn't observe any change, but in a very trivial way.

In this sense, changing the arrow of time does not simply mean taking $t\to-t$. If you were to reverse the time everywhere, both in the system's clock and in the apparatus clock, you wouldn't change anything, but in a very trivial way. This operation is equivalent to sending $i\to-i$.

The non-trivial notion of reversing the arrow of time is, sending $t\to-t$ in the system, while keeping your clock ticking in the same direction. (Or, as before, the other way around; but not both). If you do this, you can answer whether the system is invariant under time-reversal or not.

Phase transitions, mechanical damping, nuclear decay, etc. are all invariant under the trivial notion of sending $t\to-t$, meaning that you reverse the direction of time everywhere, both in the physical system and in the observer. In fact, all of physics is invariant under this operation. But these phenomena are not invariant under the physical notion of time-reversal, meaning sending $t\to-t$ in the physical system but keeping the observer's clock fixed.

The operation $i\to-i$ is equivalent to the trivial notion of reversing the arrow of time, and they are both meaningless, purely formal operations, which merely define conventions. Sending $i\to-i$ is not equivalent to the physical notion of reversing the arrow of time, the latter being a non-trivial operation which often does not leave dynamics invariant.

Answer 2

Let's use the matrix representation of complex numbers. In this representation real numbers are proportional to the unit matrix, while the imaginary unit has the form

$$i=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}.\tag1$$

Now, consider the Schrödinger's equation

$$i\partial_t |\psi\rangle-H|\psi\rangle=0.\tag2$$

Rewrite it in the following form, splitting the Hamiltonian and $|\psi\rangle$ into real and imaginary parts:

$$(i\partial_t-\Re H-i\Im H)(\Re|\psi\rangle+i\Im|\psi\rangle)=0.\tag3$$

Using $(1)$, we now can say

$$\begin{pmatrix} -\Re H& \partial_t-\Im H\\ -\partial_t+\Im H & -\Re H \end{pmatrix}\cdot \begin{pmatrix} \Re|\psi\rangle& \Im|\psi\rangle\\ -\Im|\psi\rangle & \Re|\psi\rangle \end{pmatrix}= \begin{pmatrix}0&0\\ 0&0\end{pmatrix},\tag4$$

or, expanding,

$$\begin{pmatrix} (-\partial_t+\Im H)\,\Im|\psi\rangle-\Re H\,\Re|\psi\rangle & -\Re H\,\Im|\psi\rangle+(\partial_t-\Im H)\,\Re|\psi\rangle\\ \Re H\,\Im|\psi\rangle-(\partial_t-\Im H)\,\Re|\psi\rangle & (-\partial_t+\Im H)\,\Im|\psi\rangle-\Re H\,\Re|\psi\rangle \end{pmatrix}= \begin{pmatrix}0&0\\ 0&0\end{pmatrix}.\tag5$$

What's going to happen if instead of $(1)$ we take

$$j=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix},\tag6$$

and replace $i$ with $j$ everywhere in $(3)$? We'll effectively transpose the matrices in $(4)$ and $(5)$. But this doesn't change anything at all: the equation, when considered as a system of per-component equations, remains the same. (In fact, the top row in $(5)$ is equivalent to the bottom one.)

Thus, consistent renaming $i\leftrightarrow -i$ doesn't change physics. You just have to really be consistent—not miss any instance of the imaginary unit, including the hidden ones like in the $H$ and $|\psi\rangle$ in $(2)$. Failure to be consistent will indeed lead to time reversal or some other, potentially senseless, results.

Answer 3

This is a good intuition! Yes, you can always take the complex conjugate of everything and thereby get a new theory whose predictions are the same as the old theory. In this case the waves rotate the opposite way around the complex plane, so a state with energy $E_n$ becomes a phasor $\exp({+i}E_nt/\hbar)$ rather than ${-i}.$ Arguably this is more intuitive, I feel like I had to baby-step undergraduates through the “usual” expression $\exp({-i}E_nt/\hbar)$ several times because of that minus sign.

However, that is not the only $i$ in quantum mechanics, the other main one lies in the canonical commutation relations $[\hat r_m, \hat p_n] =i\hbar\delta_{mn}$ which makes $\hat{\mathbf p}={-i}\hbar\nabla$. So we should probably also say $[\hat r_m, \hat p_n] ={-i}\hbar\delta_{mn}$ and $\hat{\mathbf p}={+i}\hbar\nabla$ as well, which again I don't terribly have a problem with.

Just to give a sense of why this $i$ also needs to change: you used to have a plane wave moving forward as $\exp({-i}\omega t + i k x),$ this is if you like the unnormalizable $\psi(x)$ that corresponds to a system with a definite momentum $\mathbf p = p\hat x$ and it is unnormalizable because of Heisenberg uncertainty: infinite momentum precision has led to infinite position imprecision. Now, that it moves forward is given by this $x-vt$ pattern where they have opposite signs for positive $\omega,k$. That pattern $x-vt$ always gives a wave traveling with velocity $+v$. And so in the traditional convention this operator $i\hbar \partial_t$ picked out the Einstein-Planck energy $E=\hbar\omega$ directly, but you had to have the opposite sign $p=-i\hbar\nabla$ to also get the positive sign on the de Broglie relation $p=\hbar k.$ in your convention you want to change the first one, but if you don't also change the second then forward-moving positive momentum waves have negative $k$ which is confusing. In this very narrow sense I suppose you could say that you then have reversed the arrow of time instead—but I think swapping the sign convention on the $\mathbf k$-vectors is way way too confusing.

But you should instead build on the plane wave $\exp({+i}\omega t - i k x),$ with your $E=-i\hbar\partial_t$ and set $p={+i}\hbar\nabla$ to avoid confusion: these canonical commutation relations really do need to change to keep sanity.

Similar $i$s exist in say the Pauli matrix $\sigma_y$ and the angular momentum expressions, one might wish to tweak those. Those are more of a consequence of the canonical commutation relations changing sign though, that is nothing “new.”