Crystal Wheeler

Answered

2022-07-12

I'm trying to solve the initial value problem $(i{\mathrm{\partial}}_{t}+{\mathrm{\Delta}}_{x})u(t,x)=0$ $u(0,x)=f(x)$ for the Schrödinger equation ($x\in {\mathbb{R}}^{n}$, f Schwartz).

I know that a fundamental solution is given by $K(t,x)=(4\pi it{)}^{-n/2}{e}^{i|x{|}^{2}/4t}$.

How do I interpret $\sqrt{i}$ here?

I'm trying to show that if I convolve the above fundamental solution K with the initial data f (convolution in the spatial variable x), then I obtain the solution to the initial value problem.

Specifically, how do I prove that $K\ast f\to f$as $t\to 0$? More generally, what are the differences between this problem and the analogous problem for the heat equation $({\mathrm{\partial}}_{t}-{\mathrm{\Delta}}_{x})u(t,x)=0$?

[I know that the Schrödinger equation and fundamental solution are obtained from their heat counterparts via $t\mapsto it$

Why is the Schrödinger equation time reversible (i.e. why can it be solved both forwards and backwards in time), while the heat equation isn't? The total integral of the heat kernel (with respect to x) is 1; is the total integral of the "Schrödinger kernel" K also equal to 1?

I know that a fundamental solution is given by $K(t,x)=(4\pi it{)}^{-n/2}{e}^{i|x{|}^{2}/4t}$.

How do I interpret $\sqrt{i}$ here?

I'm trying to show that if I convolve the above fundamental solution K with the initial data f (convolution in the spatial variable x), then I obtain the solution to the initial value problem.

Specifically, how do I prove that $K\ast f\to f$as $t\to 0$? More generally, what are the differences between this problem and the analogous problem for the heat equation $({\mathrm{\partial}}_{t}-{\mathrm{\Delta}}_{x})u(t,x)=0$?

[I know that the Schrödinger equation and fundamental solution are obtained from their heat counterparts via $t\mapsto it$

Why is the Schrödinger equation time reversible (i.e. why can it be solved both forwards and backwards in time), while the heat equation isn't? The total integral of the heat kernel (with respect to x) is 1; is the total integral of the "Schrödinger kernel" K also equal to 1?

Answer & Explanation

postojahob

Expert

2022-07-13Added 13 answers

How do I interpret $\sqrt{i}$ here?

The two roots differ by a factor of −1. For one of the roots $\underset{t\to 0}{lim}K\ast f=f$, for the other, $\underset{t\to 0}{lim}K\ast f=-f$. Choose the one that gives the former (the correct one is the one given by the standard square root on C with branch cut along the negative real axis, so $\sqrt{i}=\mathrm{exp}\pi i/4$.

Convergence as $t\to 0$.

If the data f is in Schwartz class, you can just do it via the Fourier transform. But if you want to do it in more general function spaces, and get estimates on pointwise convergence, the issue is actually amazingly delicate (and not fully resolved yet). The relevant papers are that of Sjölin and Vega written simultaneously but independently.

Difference from the heat equation.

In the heat equation case, the convolution kernels $f$ for t>0 are actually Schwartz functions, and one has that family actually form a family of "Good Kernels" (in the terminology of E. Stein) for which one actually has the general theorem that

Theorem If $f$ is an integrable function, and $f$ is continuous at ${x}_{0}$. Let ${H}_{t}$ be a family of "Good Kernels" (or approximations to the identity) then l$\underset{t\to 0}{lim}{H}_{t}\ast f({x}_{0})={x}_{0}$

The Schrödinger kernel is actually quite far from being a "Good kernel" (the criteria for which are (a) integral 1 (b) absolute integral bounded (c) restricted away from the origin, the limt→0 of the absolute integral goes to zero; (b) and (c) quite clearly fails for the Schrödinger kernel). And so the above general theorem cannot be applied.

But is the total integral 1?

Yes. While the absolute integral of the Schrödinger kernel does not converge, its improper integral converges to 1 (one can evaluate this by, for example, taking a contour integral). An indication to this is that its Fourier transform is, by definition,

$\hat{K}(t,\xi )=\mathrm{exp}(-it{\xi}^{2})\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\hat{K}(t,0)=\mathrm{exp}0=1$

Then we can note that by definition of the Fourier transform

$\hat{K}(t,\xi )={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}K(t,x)\mathrm{d}x.$

So assuming all the relevant quantities converge the value of its total integral must be 1.

(One way to make the argument above still more precise is that for any Schwarz function f one verifies, through the property of K as a Fourier multiplier, that $\int K\ast f\mathrm{d}x=\int f\mathrm{d}x$

Time reversibility?

One "meta"-argument that the Schrödinger equation, if can be solved locally, must be time reversible, lies in the form of the equation. Observe that the complex conjugation operation can be combined with the time reversal to show time reversibility (if $\mathrm{\Psi}$ is a solution to Schrödinger's equation,$\overline{\mathrm{\Psi}}$solves the complex-conjugate equation, which is the same equation as the time-reversed equation). This lies in the fact that there is no canonical choice of which square root of -1 we call i and which we call −i. So if the "Wick rotation" of the heat equation were to make sense as an evolution equation, it must be evolvable in both +i and −i imaginary time, and so must be time reversible.

Mathematically (in order to illustrate the basic intuition, I shall commit the sin of lying by omission of more difficult details), the difference between the heat and Schrödinger equations is (roughly speaking) the difference between the real and complex exponential function ${e}^{t}$ and ${e}^{it}$ for $t\in \mathbb{R}$. Up to complex conjugation, the complex exponential is "time symmetric": ${e}^{i(-t)}=\overline{{e}^{it}}$. But the function et is quite clearly not time symmetric. Now, writing the formal solution to the heat equation

${\mathrm{\partial}}_{t}u=\mathrm{\u25b3}u$

using ODE type notation

$u(t)={e}^{t\mathrm{\u25b3}}{u}_{0}$

we see that since $\mathrm{\u25b3}$ is a self adjoint operator with negative eigenvalues, ${e}^{t\mathrm{\u25b3}}$ is a contraction, and like ${e}^{t(-1)}$ is not time reversible, whereas

$i{\mathrm{\partial}}_{t}u=\mathrm{\u25b3}u$

is solved by

$u(t)={e}^{it\mathrm{\u25b3}}{u}_{0}$

and now since $\mathrm{\u25b3}$ is self-adjoint, its eigenvalues are real, we have that the exponent is purely imaginary. So the solution operator behaves like the complex exponential ${e}^{it(-1)}$ which, as discussed above, is time reversible.

The two roots differ by a factor of −1. For one of the roots $\underset{t\to 0}{lim}K\ast f=f$, for the other, $\underset{t\to 0}{lim}K\ast f=-f$. Choose the one that gives the former (the correct one is the one given by the standard square root on C with branch cut along the negative real axis, so $\sqrt{i}=\mathrm{exp}\pi i/4$.

Convergence as $t\to 0$.

If the data f is in Schwartz class, you can just do it via the Fourier transform. But if you want to do it in more general function spaces, and get estimates on pointwise convergence, the issue is actually amazingly delicate (and not fully resolved yet). The relevant papers are that of Sjölin and Vega written simultaneously but independently.

Difference from the heat equation.

In the heat equation case, the convolution kernels $f$ for t>0 are actually Schwartz functions, and one has that family actually form a family of "Good Kernels" (in the terminology of E. Stein) for which one actually has the general theorem that

Theorem If $f$ is an integrable function, and $f$ is continuous at ${x}_{0}$. Let ${H}_{t}$ be a family of "Good Kernels" (or approximations to the identity) then l$\underset{t\to 0}{lim}{H}_{t}\ast f({x}_{0})={x}_{0}$

The Schrödinger kernel is actually quite far from being a "Good kernel" (the criteria for which are (a) integral 1 (b) absolute integral bounded (c) restricted away from the origin, the limt→0 of the absolute integral goes to zero; (b) and (c) quite clearly fails for the Schrödinger kernel). And so the above general theorem cannot be applied.

But is the total integral 1?

Yes. While the absolute integral of the Schrödinger kernel does not converge, its improper integral converges to 1 (one can evaluate this by, for example, taking a contour integral). An indication to this is that its Fourier transform is, by definition,

$\hat{K}(t,\xi )=\mathrm{exp}(-it{\xi}^{2})\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\hat{K}(t,0)=\mathrm{exp}0=1$

Then we can note that by definition of the Fourier transform

$\hat{K}(t,\xi )={\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}K(t,x)\mathrm{d}x.$

So assuming all the relevant quantities converge the value of its total integral must be 1.

(One way to make the argument above still more precise is that for any Schwarz function f one verifies, through the property of K as a Fourier multiplier, that $\int K\ast f\mathrm{d}x=\int f\mathrm{d}x$

Time reversibility?

One "meta"-argument that the Schrödinger equation, if can be solved locally, must be time reversible, lies in the form of the equation. Observe that the complex conjugation operation can be combined with the time reversal to show time reversibility (if $\mathrm{\Psi}$ is a solution to Schrödinger's equation,$\overline{\mathrm{\Psi}}$solves the complex-conjugate equation, which is the same equation as the time-reversed equation). This lies in the fact that there is no canonical choice of which square root of -1 we call i and which we call −i. So if the "Wick rotation" of the heat equation were to make sense as an evolution equation, it must be evolvable in both +i and −i imaginary time, and so must be time reversible.

Mathematically (in order to illustrate the basic intuition, I shall commit the sin of lying by omission of more difficult details), the difference between the heat and Schrödinger equations is (roughly speaking) the difference between the real and complex exponential function ${e}^{t}$ and ${e}^{it}$ for $t\in \mathbb{R}$. Up to complex conjugation, the complex exponential is "time symmetric": ${e}^{i(-t)}=\overline{{e}^{it}}$. But the function et is quite clearly not time symmetric. Now, writing the formal solution to the heat equation

${\mathrm{\partial}}_{t}u=\mathrm{\u25b3}u$

using ODE type notation

$u(t)={e}^{t\mathrm{\u25b3}}{u}_{0}$

we see that since $\mathrm{\u25b3}$ is a self adjoint operator with negative eigenvalues, ${e}^{t\mathrm{\u25b3}}$ is a contraction, and like ${e}^{t(-1)}$ is not time reversible, whereas

$i{\mathrm{\partial}}_{t}u=\mathrm{\u25b3}u$

is solved by

$u(t)={e}^{it\mathrm{\u25b3}}{u}_{0}$

and now since $\mathrm{\u25b3}$ is self-adjoint, its eigenvalues are real, we have that the exponent is purely imaginary. So the solution operator behaves like the complex exponential ${e}^{it(-1)}$ which, as discussed above, is time reversible.

Most Popular Questions