In a post from a few days ago, I briefly mentioned the notion of imaginary time with regard to angular momentum. I'd like to go into that a little further in this post.

In 3 spatial dimensions, the flat (Euclidean) metric is $\eta_{ij} = \delta_{ij}$, which is quite convenient, as lengths are given by $(\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ which is just the usual Pythagorean theorem. When a temporal dimension is added, as in special relativity, the coordinates are now $x^{\mu} = (ct, x_{j})$, and the Euclidean metric becomes the Minkowski metric $\eta_{\mu \nu} = \mathrm{diag}(-1, 1, 1, 1)$ so that $\eta_{tt} = -1$, $\eta_{(t, j)} = 0$, and $\eta_{ij} = \delta_{ij}$. This means that spacetime intervals become $(\Delta s)^2 = -(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$, which is the normal Pythagorean theorem only if $\Delta t = 0$. In general, time coordinate differences contribute negatively to the spacetime interval. In addition, Lorentz transformations are given by a hyperbolic rotation by a [hyperbolic] angle $\alpha$ equal to the rapidity given by $\frac{v}{c} = \tanh(\alpha)$. This doesn't look quite the same as normal Euclidean geometry. However, a transformation to imaginary time, called a Wick rotation, can be done by setting $\tau = it$, so $x^{\mu} = (ic\tau, x_{j})$, $\eta_{\mu \nu} = \delta_{\mu \nu}$, $(\Delta s)^2 = (c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ as in the usual Pythagorean theorem, and the Lorentz transformation is given by a real rotation by an angle $\theta = i\alpha$ (though I may have gotten some of these signs wrong so forgive me) where $\alpha$ is now imaginary. Now, the connection to the component $L_{(0, j)}$ of the angular momentum tensor should be more clear.

I first encountered this in the class 8.033 — Relativity, where I was able to explore this curiosity on a problem set. That question and the accompanying discussion seemed to say that while this is a cool thing to try doing once, it isn't really useful, especially because it does not hold true in general relativity with more general metrics $g_{\mu \nu} \neq \eta_{\mu \nu}$ except in very special cases. However, as it turns out, imaginary time does play a role in quantum mechanics, even without the help of relativity.

SchrÃ¶dinger time evolution occurs through the unitary transformation $u = e^{-\frac{itH}{\hbar}}$ satisfying $uu^{\dagger} = u^{\dagger} u = 1$. This means that the probability that an initial state $|\psi\rangle$ ends after time $t$ in the same state is given by the amplitude (whose square is the probability [density]) $\mathfrak{p}(t) = \langle\psi|e^{-\frac{itH}{\hbar}}|\psi\rangle$. Meanwhile, assuming the states $|\psi\rangle$ form a complete and orthonormal basis (though I don't know if this assumption is truly necessary), the partition function $Z = \mathrm{trace}\left(e^{-\frac{H}{k_B T}}\right)$, which can be expanded in the basis $|\psi\rangle$ as $Z = \sum_{\psi} \langle\psi|e^{-\frac{H}{k_B T}}|\psi\rangle$. This, however, is just as well rewritten as $Z = \sum_{\psi} \mathfrak{p}\left(t = -\frac{i\hbar}{k_B T}\right)$. Hence, quantum and statistical mechanical information can be gotten from the same amplitudes using the substitution $t = -\frac{i\hbar}{k_B T}$, which essentially calls temperature a reciprocal imaginary time. This is not really meant to show anything more deep or profound about the connection between time and temperature; it is really more of a trick stemming from the fact that the same Hamiltonian can be used to solve problem in quantum mechanics or equilibrium statistical mechanics.

As an aside, it turns out that temperature, even when measured in an absolute scale, can be negative. There are plenty of papers of this online, but suffice it to say that this comes from a more general statistical definition of temperature. Rather than defining it (as it commonly is) as the average kinetic energy of particles, it is better to define it as a measure of the probability distribution that a particle will have a given energy. Usually, particles tend to be in lower energy states more than in higher energy states, and as a consequence, the temperature is positive. However, it is possible (and has been done repeatedly) under certain circumstances to cleverly force the system in a way that causes particles to be in higher energy states with higher probability than in lower energy states, and this is exactly the negative temperature. More formally, $\frac{1}{T} = \frac{\partial S}{\partial E}$ where $E$ is the energy and $S$ is the entropy of the system, which is a measure of how many different states the system can possibly have for a given energy. For positive temperature, if two objects of different temperatures are brought into contact, energy will flow from the hotter one to the colder, cooling the former and heating the latter until equal temperatures are achieved. For negative temperature, though, if an object with negative temperature is brought in contact with an object that has positive temperature, each object tends to increase its own entropy. Like most normal objects, the latter does this by absorbing energy, but by the definition of temperature, the former does this by releasing energy, meaning the former will spontaneously heat the latter. Hence, negative temperature is hotter than positive temperature; this is a quirk of the definition of reciprocal temperature, so really what is happening is that absolute zero on the positive side is still the coldest possible temperature, absolute zero on the negative side is now the hottest temperature, and $\pm \infty$ is in the middle.

This was really just me writing down stuff that I had been thinking about a couple of months ago. I hope this helps someone, and I also await the day when TV newscasters say "complex time brought to you by..." instead of "time and temperature brought to you by...".

In 3 spatial dimensions, the flat (Euclidean) metric is $\eta_{ij} = \delta_{ij}$, which is quite convenient, as lengths are given by $(\Delta s)^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ which is just the usual Pythagorean theorem. When a temporal dimension is added, as in special relativity, the coordinates are now $x^{\mu} = (ct, x_{j})$, and the Euclidean metric becomes the Minkowski metric $\eta_{\mu \nu} = \mathrm{diag}(-1, 1, 1, 1)$ so that $\eta_{tt} = -1$, $\eta_{(t, j)} = 0$, and $\eta_{ij} = \delta_{ij}$. This means that spacetime intervals become $(\Delta s)^2 = -(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$, which is the normal Pythagorean theorem only if $\Delta t = 0$. In general, time coordinate differences contribute negatively to the spacetime interval. In addition, Lorentz transformations are given by a hyperbolic rotation by a [hyperbolic] angle $\alpha$ equal to the rapidity given by $\frac{v}{c} = \tanh(\alpha)$. This doesn't look quite the same as normal Euclidean geometry. However, a transformation to imaginary time, called a Wick rotation, can be done by setting $\tau = it$, so $x^{\mu} = (ic\tau, x_{j})$, $\eta_{\mu \nu} = \delta_{\mu \nu}$, $(\Delta s)^2 = (c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2$ as in the usual Pythagorean theorem, and the Lorentz transformation is given by a real rotation by an angle $\theta = i\alpha$ (though I may have gotten some of these signs wrong so forgive me) where $\alpha$ is now imaginary. Now, the connection to the component $L_{(0, j)}$ of the angular momentum tensor should be more clear.

I first encountered this in the class 8.033 — Relativity, where I was able to explore this curiosity on a problem set. That question and the accompanying discussion seemed to say that while this is a cool thing to try doing once, it isn't really useful, especially because it does not hold true in general relativity with more general metrics $g_{\mu \nu} \neq \eta_{\mu \nu}$ except in very special cases. However, as it turns out, imaginary time does play a role in quantum mechanics, even without the help of relativity.

SchrÃ¶dinger time evolution occurs through the unitary transformation $u = e^{-\frac{itH}{\hbar}}$ satisfying $uu^{\dagger} = u^{\dagger} u = 1$. This means that the probability that an initial state $|\psi\rangle$ ends after time $t$ in the same state is given by the amplitude (whose square is the probability [density]) $\mathfrak{p}(t) = \langle\psi|e^{-\frac{itH}{\hbar}}|\psi\rangle$. Meanwhile, assuming the states $|\psi\rangle$ form a complete and orthonormal basis (though I don't know if this assumption is truly necessary), the partition function $Z = \mathrm{trace}\left(e^{-\frac{H}{k_B T}}\right)$, which can be expanded in the basis $|\psi\rangle$ as $Z = \sum_{\psi} \langle\psi|e^{-\frac{H}{k_B T}}|\psi\rangle$. This, however, is just as well rewritten as $Z = \sum_{\psi} \mathfrak{p}\left(t = -\frac{i\hbar}{k_B T}\right)$. Hence, quantum and statistical mechanical information can be gotten from the same amplitudes using the substitution $t = -\frac{i\hbar}{k_B T}$, which essentially calls temperature a reciprocal imaginary time. This is not really meant to show anything more deep or profound about the connection between time and temperature; it is really more of a trick stemming from the fact that the same Hamiltonian can be used to solve problem in quantum mechanics or equilibrium statistical mechanics.

As an aside, it turns out that temperature, even when measured in an absolute scale, can be negative. There are plenty of papers of this online, but suffice it to say that this comes from a more general statistical definition of temperature. Rather than defining it (as it commonly is) as the average kinetic energy of particles, it is better to define it as a measure of the probability distribution that a particle will have a given energy. Usually, particles tend to be in lower energy states more than in higher energy states, and as a consequence, the temperature is positive. However, it is possible (and has been done repeatedly) under certain circumstances to cleverly force the system in a way that causes particles to be in higher energy states with higher probability than in lower energy states, and this is exactly the negative temperature. More formally, $\frac{1}{T} = \frac{\partial S}{\partial E}$ where $E$ is the energy and $S$ is the entropy of the system, which is a measure of how many different states the system can possibly have for a given energy. For positive temperature, if two objects of different temperatures are brought into contact, energy will flow from the hotter one to the colder, cooling the former and heating the latter until equal temperatures are achieved. For negative temperature, though, if an object with negative temperature is brought in contact with an object that has positive temperature, each object tends to increase its own entropy. Like most normal objects, the latter does this by absorbing energy, but by the definition of temperature, the former does this by releasing energy, meaning the former will spontaneously heat the latter. Hence, negative temperature is hotter than positive temperature; this is a quirk of the definition of reciprocal temperature, so really what is happening is that absolute zero on the positive side is still the coldest possible temperature, absolute zero on the negative side is now the hottest temperature, and $\pm \infty$ is in the middle.

This was really just me writing down stuff that I had been thinking about a couple of months ago. I hope this helps someone, and I also await the day when TV newscasters say "complex time brought to you by..." instead of "time and temperature brought to you by...".

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