# Solving PDE by means of ODE

1. Sep 26, 2011

### giova7_89

Here's my question: as soon as I learned Quantum Mechanics and Schrodinger equation, I saw a "similarity" with the equation one gets in classical mechanics for the evolution of a function in phase space. In QM one has:

$i\hbar\frac{d}{dt}\psi = \hat{H}\psi$

and this is a evolution equation where $\psi$ is the element which evolves and it is an element of a space of functions.

If one represents this equation (considering one spinless particle) in the $|\vec{x}>$ basis, one gets the wave equation that everyone knows, where the hamiltonian on this basis acts on the state ket as

$-\frac{\hbar^2}{2m}\nabla^2 + U(\vec{x})$

does.

In CM one has:

$\frac{d}{dt}f = \hat{L}f$

where f is the element that evolves and it is an element of a space on functions, too. (here I assumed that the functions I want to evolve from time t0 to time t do not depend on t0 explicitly, otherwise I should have added $\partial_t f$ to that equation)

If one represents that equation in the $|\vec{q},\vec{p}>$ basis one gets:

$\frac{d}{dt}f (\vec{q},\vec{p}) = \{f(\vec{q},\vec{p}),H(\vec{q},\vec{p})\}$

and if I solve Hamilton equations and get the hamiltonian flow $\Phi^H_{(t,t_0)}$, I know that the solution to the equation with initial condition f0 is:

$(e^{\hat{L}\Delta t}[f])(\vec{q},\vec{p}) = f(\Phi^H_{(t,t_0)}(\vec{q},\vec{p}))$

(I assumed that H does not depend on time).

Then my question is: in CM i can solve the evolution equation for f (a PDE) by solving ODEs. Can a similar thing be done in QM with Schrodinger equation? Is there any vector field $\vec{X}$ whose associated flow (which i can find by solving $\frac{d}{dt}\vec{x} = \vec{X}$) one can use to evolve the initial state ket of QM?

2. Sep 26, 2011

### Ben Niehoff

If the Hamiltonian does not depend on time, one can write the solution formally as

$$\lvert \psi (t) \rangle = e^{\frac{\Delta t}{i \hbar} \hat H} \lvert \psi (0) \rangle$$
I think it's correct to say this is some kind of flow on state space. Typically, the flow will be unitary; i.e. preserves inner products.

If the Hamiltonian does depend on time, then the time evolution operator will be a "path ordered exponential"

$$\mathcal P \exp \int_0^t \hat H(t') dt'$$

3. Sep 26, 2011

### giova7_89

But how can I find the vector field that generates this flow? I tried (naively) to get the three "equations of motion" by chosing as $\psi(x,y,z)$ the functions $\psi(x,y,z) = x$ , $\psi(x,y,z) = y$ , $\psi(x,y,z) = z$. By doing this I got that:

$\frac{d}{dt}\vec{r} = \frac{1}{i\hbar}U(\vec{r})\vec{r}$

I tried to solve this in some simple cases but I got no agreement with the actual solution one obtains by using the quantum evolution operator $\hat{U}_ {t,t_0} = e^{\frac{\hat{H}\Delta t}{i\hbar}}$

4. Sep 26, 2011

### Ben Niehoff

The state is a vector in an infinite-dimensional Hilbert space; not a vector in R^3.

5. Sep 26, 2011

### giova7_89

I know.. But in classical mechanics an observable is an element of a infinite dimensional space too! The space of functions of position and momentum. Maybe I didn't made myself clear enough in my last post: from what i understood about CM and its equations, one can find the hamiltonian flow by solving ODEs and plug it in the initial "phase function" f0 once it has been represented in $|\vec{q},\vec{p}>$ to get the "phase function " at time t and solve equations which involve the liouville operator.

This way one solves the equation

$\frac{d}{dt}f = \hat{L}f$

where $\hat{L}$ is the Liouville operator

Representing functions of $|\vec{q},\vec{p}>$ is similar as representing quantum states in the $|\vec{x}>$ basis!

I wanted to know if a similar procedure can be obtained for the schrodinger equation (and, in general, for other kinds of "initial value PDEs"). I mean: is there an equation

$\frac{d}{dt}\vec{x} = \vec{X}(\vec{x})$

which once solved gives

$\vec{x}(t)$

such that

$(\hat{U}_{(t,t_0)}[\psi_{t_0}])(\vec{x}) = \psi_{t_0}(\vec{x}(t))$
(where $\hat{U}_{(t,t_0)}$ is the evolution operator of QM)??

I hope I made myself clear!!

PS: thanks a lot for your replies!

Last edited: Sep 26, 2011
6. Sep 27, 2011

### spocchio

I do not think that all the Hamiltonians can accommodate your request that you can find a $\frac{d}{dt}\vec{x} = \vec{X}$ for each $\hat{H}$ .

But I'll show you a class of Hamiltonians wherewith you're able to do it, unlucky these Hamiltonians don't have a quantum physical sense, but it's stll mathematically correct.

Let's consider an Hamiltonian in the form:
$\hat{H}=f_2(\hat{q_2})\hat{p_1}-f_1(\hat{q_1})\hat{p_2}$.

With $\left[\hat{q_i},\hat{p_i}\right]=i\hbar$
and the evolution under the well known Schrodingher equation $i\hbar\frac{\partial}{\partial t}\lvert\psi_t\rangle = \hat{H}\lvert\psi_t\rangle$.

Now see what equation we obtain in the $\lvert q_1,q_2\rangle$ base:

$f_2(q_2)\frac{\partial \psi_t(q_1,q_2) }{\partial q_1}-f_1(q_1)\frac{\partial \psi_t(q_1,q_2) }{\partial q_2} = -\frac{\partial}{\partial t}\psi_t(q_1,q_2)$

and we can always find an H(q1,q2) that satisfies

$f_1(q)=\frac{\partial H(q1,q2)}{\partial q_1}$
$f_2(q)=\frac{\partial H(q1,q2)}{\partial q_2}$.

The first equation of psi became:

$\frac{\partial H(q1,q2)}{\partial q_2}\frac{\partial \psi_t(q_1,q_2) }{\partial q_1}-\frac{\partial H(q1,q2)}{\partial q_1}\frac{\partial \psi_t(q_1,q_2) }{\partial q_2} = -\frac{\partial}{\partial t}\psi_t(q_1,q_2)$

That's formally identical to the equation $\frac{\partial}{\partial t} \psi= \hat{L}\psi$ under this Hamiltonian flux!

Last edited: Sep 27, 2011
7. Sep 27, 2011

### spocchio

let's take another example,this time we'll consider a well known quantum problem and it's related Hamiltonian.
Consider a quantum particle in free fall:

$\hat{H}=\frac{1}{2m}\hat{p}^2+mg\hat{q}$

let's see the scrodingher equation in$\lvert p \rangle$

$-\frac{p^2}{2mi\hbar}\psi(p,t)-mg\frac{\partial}{\partial p}\psi(p,t)+\frac{\partial}{\partial t}\psi(p,t) = 0$

now take this equation $\frac{d}{dt}\vec{x} = \vec{X}$ where $\vec{x} \in R^1$ with $\vec{x} = (p)$; and $\vec{X}=(-mg)$

and so the schrodingher equation became

$-\frac{p^2}{2mi\hbar}\psi(p,t)+\frac{d}{dt}\psi(p,t) = 0$

under the flux $\frac{d}{dt}p = -mg$

as you can see, this trick is possible due to the linearity of partial derivates in the scrodingher equation.

Last edited: Sep 27, 2011
8. Sep 27, 2011

### giova7_89

Hi spocchio!

I didn't get what you were trying to say in your last post, but I tried to solve that problem.

First off I took as initial state ket the function $\psi (p) = p^2$ and made it evolve: i took terms not higher than order $t$ and got (I set $\hbar = 1$ )

$(\hat{U}_t\psi)(p) = p^2 - 2 g m p t - \frac{i t p^4}{m} + O(t^2)$

Then I chose as initial function $\phi (p) = p$ and evolved it: I got $p(t)$ by doing so. Then i calculated $\psi (p (t))$ and got

$\psi(p(t)) = p^2 - 2 g m p t - \frac{i t p^4}{2m} + O(t^2)$

which is different from the actual solution!!! This means one cannot just plug the "coordinate function" (here this function is $\phi (p) = p$) in the Schrodinger equation, solve the ODE that comes out and then put the solution inside the original function to get the solution to the initial value problem one he started with.

In classical mechanics, however, that works: that is the two formulas i wrote up there are equal.

Going back to your post: I don't understand the meaning of $\frac{d}{dt}p = -mg$ in this contest.

If I use the equation $\frac{d}{dt}p = -mg$ you suggested, I obtain $p(t) = p -mgt$ and $\psi(p(t)) = (p-mgt)^2$ isn't equal to the solution to the actual Scrhodinger equation with initial ket $|\psi>$...

9. Sep 27, 2011

### spocchio

The solution are the same under your approimation:
since $\frac{d}{dt}p=-mg$ you have that $dp \cong dt$
and so, in an approximation of $O(t^2)$ to check the solution you have to discard $O(p^2)$

$\psi(p(t)) = p^2 - 2 g m p t - \frac{i t p^4}{2m} + O(t^2) = - 2 g m p t + O(t^2) + O(p^2)$
$\psi(p(t)) = (p-mgt)^2 = -2mgpt + O(t^2) + O(p^2)$

But,ok, I agree my example have some issues, but i dont get where, if i have

$-\frac{p^2}{2mi\hbar}\psi(p,t)-mg \frac{\partial}{\partial p}\psi(p,t)+\frac{\partial}{\partial t}\psi(p,t) = 0$

and set $\dot{p}=-mg$
and ask (can't I?) $\frac{d}{dt}\psi(p,t)=\dot{p} \frac{\partial}{\partial p}\psi(p,t)+\frac{\partial}{\partial t}\psi(p,t)$

and so i obtain

$-\frac{p^2}{2mi\hbar} \psi(p,t)+\dot{p} \frac{\partial}{\partial p}\psi(p,t)+\frac{\partial}{\partial t}\psi(p,t) = 0$

that became

$-\frac{p^2}{2mi\hbar}\psi(p,t)+\frac{d}{dt}\psi(p,t) = 0$

where is my error?

Last edited: Sep 27, 2011
10. Sep 27, 2011

### giova7_89

No the example is ok! I did something similar for the free particle.. The thing is that it doesn't answer my problem.. I am looking for a vector field whose flow I can use to solve EXACTLY a PDE evolution equation, as I can do in classical mechanics.

The main point of my last post was, in fact, that

$(\hat{U}_t\psi)(p) \neq \psi(p(t))$

while in CM this holds true..

The equation I refer to is one of the last equations of my first post.

I was just wondering if one can do that for all evolution equations, and if so how one does that..

Last edited: Sep 27, 2011