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Consider the free Schrödinger equation in ${d}$ spatial dimensions, which I will normalise as

$\displaystyle i u_t + \frac{1}{2} \Delta_{{\bf R}^d} u = 0 \ \ \ \ \ (1)$

where ${u: {\bf R} \times {\bf R}^d \rightarrow {\bf C}}$ is the unknown field and ${\Delta_{{\bf R}^{d+1}} = \sum_{j=1}^d \frac{\partial^2}{\partial x_j^2}}$ is the spatial Laplacian. To avoid irrelevant technical issues I will restrict attention to smooth (classical) solutions to this equation, and will work locally in spacetime avoiding issues of decay at infinity (or at other singularities); I will also avoid issues involving branch cuts of functions such as ${t^{d/2}}$ (if one wishes, one can restrict ${d}$ to be even in order to safely ignore all branch cut issues). The space of solutions to (1) enjoys a number of symmetries. A particularly non-obvious symmetry is the pseudoconformal symmetry: if ${u}$ solves (1), then the pseudoconformal solution ${pc(u): {\bf R} \times {\bf R}^d \rightarrow {\bf C}}$ defined by

$\displaystyle pc(u)(t,x) := \frac{1}{(it)^{d/2}} \overline{u(\frac{1}{t}, \frac{x}{t})} e^{i|x|^2/2t} \ \ \ \ \ (2)$

for ${t \neq 0}$ can be seen after some computation to also solve (1). (If ${u}$ has suitable decay at spatial infinity and one chooses a suitable branch cut for ${(it)^{d/2}}$, one can extend ${pc(u)}$ continuously to the ${t=0}$ spatial slice, whereupon it becomes essentially the spatial Fourier transform of ${u(0,\cdot)}$, but we will not need this fact for the current discussion.)

An analogous symmetry exists for the free wave equation in ${d+1}$ spatial dimensions, which I will write as

$\displaystyle u_{tt} - \Delta_{{\bf R}^{d+1}} u = 0 \ \ \ \ \ (3)$

where ${u: {\bf R} \times {\bf R}^{d+1} \rightarrow {\bf C}}$ is the unknown field. In analogy to pseudoconformal symmetry, we have conformal symmetry: if ${u: {\bf R} \times {\bf R}^{d+1} \rightarrow {\bf C}}$ solves (3), then the function ${conf(u): {\bf R} \times {\bf R}^{d+1} \rightarrow {\bf C}}$, defined in the interior ${\{ (t,x): |x| < |t| \}}$ of the light cone by the formula

$\displaystyle conf(u)(t,x) := (t^2-|x|^2)^{-d/2} u( \frac{t}{t^2-|x|^2}, \frac{x}{t^2-|x|^2} ), \ \ \ \ \ (4)$

also solves (3).

There are also some direct links between the Schrödinger equation in ${d}$ dimensions and the wave equation in ${d+1}$ dimensions. This can be easily seen on the spacetime Fourier side: solutions to (1) have spacetime Fourier transform (formally) supported on a ${d}$-dimensional hyperboloid, while solutions to (3) have spacetime Fourier transform formally supported on a ${d+1}$-dimensional cone. To link the two, one then observes that the ${d}$-dimensional hyperboloid can be viewed as a conic section (i.e. hyperplane slice) of the ${d+1}$-dimensional cone. In physical space, this link is manifested as follows: if ${u: {\bf R} \times {\bf R}^d \rightarrow {\bf C}}$ solves (1), then the function ${\iota_{1}(u): {\bf R} \times {\bf R}^{d+1} \rightarrow {\bf C}}$ defined by

$\displaystyle \iota_{1}(u)(t,x_1,\ldots,x_{d+1}) := e^{-i(t+x_{d+1})} u( \frac{t-x_{d+1}}{2}, x_1,\ldots,x_d)$

solves (3). More generally, for any non-zero scaling parameter ${\lambda}$, the function ${\iota_{\lambda}(u): {\bf R} \times {\bf R}^{d+1} \rightarrow {\bf C}}$ defined by

$\displaystyle \iota_{\lambda}(u)(t,x_1,\ldots,x_{d+1}) :=$

$\displaystyle \lambda^{d/2} e^{-i\lambda(t+x_{d+1})} u( \lambda \frac{t-x_{d+1}}{2}, \lambda x_1,\ldots,\lambda x_d) \ \ \ \ \ (5)$

solves (3).

As an “extra challenge” posed in an exercise in one of my books (Exercise 2.28, to be precise), I asked the reader to use the embeddings ${\iota_1}$ (or more generally ${\iota_\lambda}$) to explicitly connect together the pseudoconformal transformation ${pc}$ and the conformal transformation ${conf}$. It turns out that this connection is a little bit unusual, with the “obvious” guess (namely, that the embeddings ${\iota_\lambda}$ intertwine ${pc}$ and ${conf}$) being incorrect, and as such this particular task was perhaps too difficult even for a challenge question. I’ve been asked a couple times to provide the connection more explicitly, so I will do so below the fold.

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