Is there the equivalent of the Schrodinger equation $i\hbar\frac{d}{dt} \psi(t) =H \psi(t)$, valid for $U=e^{-itH}$, but for a Lie group of multiple parameters.

For SU(2), one would have $U=e^{-i(x\sigma_x+y\sigma_y+z\sigma_z)}$. Then, $H=x\sigma_x+y\sigma_y+z\sigma_z$. Consider the system of three differential equations which is the "equivalent" of the Schrodinger equation but now for three parameters instead of one:

$$ i\hbar \frac{d}{dx}\psi(x,y,z)=\sigma_x \psi(x,y,z)\\ i\hbar \frac{d}{dy}\psi(x,y,z)=\sigma_y \psi(x,y,z)\\ i\hbar \frac{d}{dz}\psi(x,y,z)=\sigma_z \psi(x,y,z) $$

I am now curious; in the case of $e^{-itH}$ represents the evolution of the wave function in time. Here, what would $d/dx, d/dy, d/dz$ represent?


2 Answers 2


You're thinking about this exactly the wrong way around: The equations for one-parameter groups other than time evolution are not "equivalents of the Schrödinger equation", it is rather the case that the Schrödinger equation is the special case of the general relationship between a one-parameter group and its generator when the one-parameter group is time-evolution.

Explicitly: Stone's theorem says that every strongly continuous one-parameter group $(U_\epsilon)_{\epsilon\in\mathbb{R}}$ has an associated self-adjoint generator $T$ such that $U_\epsilon = \mathrm{e}^{\mathrm{i}T\epsilon}$. It follows directly that, for any state $\psi$, $\psi(\epsilon) = U_\epsilon \psi$ fulfills the differential equation $$ \partial_\epsilon \psi = \mathrm{i}T\psi.$$

Now, when the $U_\epsilon$ represent evolution in time, then we write $t$ for the generic parameter $\epsilon$ and $H = -T$ for the generator, and that's the Schrödinger equation.

For e.g. $\mathfrak{so}(3)$ rotations, we of course then get the one-parameter group of rotations $U_\alpha = \mathrm{e}^{\alpha\hat{n}\cdot\vec \sigma}$around an axis $\hat{n}$ (a unit vector) and the corresponding differential equation $$ \partial_\alpha \psi = \mathrm{i}\hat{n}\cdot\vec{\sigma} \psi,$$ that tells you how $\psi(\alpha) = U_\alpha\psi$ changes when the angle of the rotation $\alpha$ changes. For $\hat{n}$ the unit vector in the $x$ resp. $y$ resp. $z$ directions, this is the three equations in the question.

This equation tells you nothing more or less than how a state rotates around the axis $\hat{n}$ - solving the differential equation for $\psi(\alpha)$ gives you the same information as computing $\psi(\alpha)$ from its definition as $U_\alpha \psi$.

It's a bit more rare that a) we're actually interested in $\psi(\epsilon)$ for cases where $\epsilon$ is not time and that b) it is easier to compute the solution to the differential equation than to just directly compute the action of $U_\epsilon$ on a given $\psi$, so you don't see these "non-Schrödinger" infinitesimal versions of Stone's theorem all that often, but there isn't really anything special about them.


Let's take your $SU(2)$ example. A general rotation $U$ will act on a state $\psi$ by transforming it to a new state $\psi'$ as follows \begin{equation} \psi' = U \psi \end{equation} Now consider for simplicity an rotation about the $z$ axis by an angle $\theta$. For a general rotation about the $z$ axis, we can express $U$ as $U=e^{-i \sigma_z \theta}$. For an infintesimal angle $\delta \theta$, we can Taylor expand $U$ and obtain \begin{equation} \psi' = \left(1 - i \sigma_z \delta \theta + \cdots\right) \psi \end{equation} Rearranging this, we get the equation \begin{equation} \frac{\psi' - \psi}{\delta \theta} = -i \sigma_z \psi \end{equation} Taking the limit $\delta \theta\rightarrow 0$, we indeed get your "Schrodinger equation" \begin{equation} i \frac{d\psi}{d\theta} = \sigma_z \psi \end{equation} Or in terms of the angular momentum operator $L_z=\hbar \sigma_z$, \begin{equation} i \hbar \frac{d\psi}{d\theta} = L_z \psi \end{equation} In general, the generators of the Lie algebra tell us how to do infinitesimal transformations on the states, which we can express as a variation (or a derivative).

This is an example of a common trick that is used a lot in field theory: converting an "global" statement that works for any value of parameters (like $\psi'=U\psi$) into a "local" differential statement, by evaluating the global statement for a small variation of the parameter and Taylor expanding. Often, the local statement (a differential equation, or a variational statement) is easier to work with than the global one.

  • $\begingroup$ Thank you for the answer. Now can I ask you to consider the SO(3,1) or the Spin(3,1) group. In either of these case we would get 6 (or I believe 7 for Spin(3,1)). Using these 6 (or 7) equations, can you do something useful with them? Perhaps recover the Dirac equation (or maybe that is too ambitious)? For the spin group, these equations should be all the ways which the Dirac wave function can evolve into, no? $\endgroup$
    – Anon21
    Jul 7, 2022 at 1:06
  • $\begingroup$ @Anon21 Just FYI there were some errors in my original answer, which I corrected. In terms of your follow up question, I have to think about it. I think the answer is that you should be able to derive the Dirac equation using properties of the Poincaire algebra (Lorentz boosts + translations). For example, for a scalar field, you can derive the Klein Gordon equation $E^2-p^2=m^2$ from the algebra, since $P_\mu P^\mu = -E^2+p^2 = - m^2$ is a Casimir invariant. But, I'm not sure whether the method you're proposing is the way to go. $\endgroup$
    – Andrew
    Jul 7, 2022 at 1:14
  • $\begingroup$ If you have access to Zee's QFT book or Volume 1 of Weinberg's QFT book, I am sure this question is answered there. $\endgroup$
    – Andrew
    Jul 7, 2022 at 1:16

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