Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ordinarty differential equations

  1. Sep 16, 2014 #1
    With
    $$
    H = \alpha \left(x^{2}\partial_{x} - x \partial_{x} \right) + \beta\left(x \partial_{x}^{2} - x^{2}\partial_{x}^{2}\right)
    $$

    Sorry for the messed up tex. I don't know how to fix it. It works in my editor.

    Hallo PF! This might become a somewhat long post so for the tldr-people just jump to the bold "introduction over".

    John Baez and Jacob Biamonte recently published some lecture notes called A Course on Quantum Techniques for Stochastic Mechanics(http://arxiv.org/abs/1209.3632). In these notes they develope some techniques to investigate some ordinarty differential equations in continuus variables using a discrete description of the system instead( The master equation). It is sort of like going from the concentration of a thing to describing the actual things one by one. The basic idear is to describe the system using a state function
    $$
    \Psi(t) = \sum_{n} \psi_{n}(t)x^{n}
    $$
    ## \psi_{n} ## is the probability of having a state with n things. That is we have
    $$
    \sum_{n} \psi_{n}(t) = 1
    $$
    A pure state of n things is thus given by ## \Psi(t) = x^{n} ##. The time evolution for this system is described by a hamiltonian and is given by
    $$
    d_{t} \Psi(t) = H\Psi(t)
    $$
    Hence the thread title :). Within this framework the anahilation and creation operators is simply
    $$
    a^{\dagger} = x \quad \text{and} \quad a = \partial_{x}
    $$
    As in quantum mechanics we define an expectation value as
    $$
    < O > = \left(O \Psi\right)\big|_{z=1} = \sum O \Psi
    $$
    Where the last equality is just notation such we match the one used in the notes. The last thing we need is the number operator given by
    $$
    N = a^{\dagger} a
    $$
    As far as I have understood in principle when taking the expectation value for the number operator we should in some sense get back to the solution of the original ordinary differential equations in the continuum variable( the concentration). These things and more can be found in chapter 2-5 of the lecture notes. Chapter 6 deals with feynman diagrams for the master equation. Essentially the develope a diagramatic expansion for the master equation by expanding the solution of the master equation as
    $$
    \exp(tH)\Psi(0) = (1 +tH + \frac{1}{2}t^{2}H^{2}+ \cdots)\Psi(0)
    $$
    Unfortunetly they do not show an example where this series converge or is easy to sum. This is more or less where you, physics forum comes in :) !
    Introduction over
    The simple ordinary differential equation
    $$
    \frac{dP}{dt} = \alpha P - \beta P^{2}
    $$
    Has the solution
    $$
    P(t) = \frac{Q}{1 + A \exp(-\alpha t)} \qquad \text{with} \qquad A = \frac{Q - P(0)}{P(0)}, \qquad Q = \frac{\alpha}{\beta}
    $$
    This is the example they study in chapter 6. Using the rules developed in chapter 5 we find that this system can be described in the discrete version by the hamiltonian
    $$
    H = \alpha \left(x^{2}\partial_{x} - x \partial_{x} \right) + \beta\left(x \partial_{x}^{2} - x^{2}\partial_{x}^{2}\right)
    $$
    (See chapter 6.3). Now I would like to see if it is possible to solve the master equation perturbatively and calculating ## < N > ## to get a better and better approximation to the solution given above.

    Solution attempt 1.

    Lets just calculate the first few terms in the expansion.
    [tex]
    \begin{align}
    H^{2} =& (\beta^{2}a^{\dagger 4}-2\beta^{2}a^{\dagger3}+\beta^{2}a^{\dagger2})a^{4} \notag \\
    &+(-2\alpha\beta a^{\dagger 4}+(4\alpha\beta+4\beta^{2})a^{\dagger3}+(-2\alpha\beta-6\beta^{2})a^{\dagger2}+2\beta^{2}a^{\dagger})a^{3} \notag \\
    &+(\alpha^{2}a^{\dagger 4}+(-2\alpha^{2}-6\alpha\beta)a^{\dagger 3}+(\alpha^{2}+9\alpha\beta+2\beta^{2})a^{\dagger2}+(-3\alpha\beta-2\beta^{2})a^{\dagger})a^{2} \notag \\
    &+(2\alpha^{2}a^{\dagger 3}+(-3\alpha^{2}-2\alpha\beta)a^{\dagger2}+(\alpha^{2}+2\alpha\beta)a^{\dagger})a
    \end{align}
    [/tex]
    The next
    [tex]
    \begin{align}
    H^{3} = & (-\beta^{3}a^{\dagger6}+3\beta^{3}a^{\dagger5}-3\beta^{3}a^{\dagger4}+\beta^{3}a^{\dagger3})a^{6} \notag \\
    & +((-9\alpha\beta^{2}-12\beta^{3})a^{\dagger5}+6\beta^{3}a^{\dagger2}+(-3\alpha\beta^{2}-24\beta^{3})a^{\dagger3}+3\alpha\beta^{2}a^{\dagger6}+(9\alpha\beta^{2}+30\beta^{3})a^{\dagger4})a^{5} \notag \\ & +\big[(-9\alpha^{2}\beta-75\alpha\beta^{2}-38\beta^{3})a^{\dagger4}+(9\alpha^{2}\beta+30\alpha\beta^{2})a^{\dagger5}+6\beta^{3}a^{\dagger}-3\alpha^{2}\beta a^{\dagger6}+(-15\alpha\beta^{2}-44\beta^{3})a^{\dagger2} \notag \\ & +(3\alpha^{2}\beta+60\alpha\beta^{2}+76\beta^{3})a^{\dagger3}\big]a^{4} \notag \\
    & +\big[(3\alpha^{3}+60\alpha^{2}\beta+76\alpha\beta^{2})a^{\dagger4}+(-3\alpha^{3}-24\alpha^{2}\beta)a^{\dagger5}+(-12\alpha\beta^{2}-16\beta^{3})a^{\dagger} \notag \\
    &+\alpha^{3}a^{\dagger6}+(12\alpha^{2}\beta+88\alpha\beta^{2}+48\beta^{3})a^{\dagger2}+(-\alpha^{3}-48\alpha^{2}\beta-152\alpha\beta^{2}-32\beta^{3})a^{\dagger3}\big]a^{3} \notag \\
    & +\big[(12\alpha^{3}+88\alpha^{2}\beta+48\alpha\beta^{2})a^{\dagger3}+(7\alpha^{2}\beta+24\alpha\beta^{2}+4\beta^{3})a^{\dagger}+(-15\alpha^{3}-44\alpha^{2}\beta)a^{\dagger4}+6\alpha^{3}a^{\dagger5} \notag \\
    &+(-3\alpha^{3}-51\alpha^{2}\beta-72\alpha\beta^{2}-4\beta^{3})a^{\dagger2}\big]a^{2} \notag \\
    & +((-\alpha^{3}-8\alpha^{2}\beta-4\alpha\beta^{2})a^{\dagger}+6\alpha^{3}a^{\dagger4}+(-12\alpha^{3}-16\alpha^{2}\beta)a^{\dagger3}+(7\alpha^{3}+24\alpha^{2}\beta+4\alpha \beta^{2})a^{\dagger2})a
    \end{align}
    [/tex]
    And I have the next one as well but that probably wont help much. Looking for some thing more manageble lets look at the specific example of
    $$
    \alpha = \beta = 1 \quad \Psi(0) = x^{2}
    $$
    That is the pure state with 2 things which we will expect decays to x. Inserting I find
    [tex]
    \begin{align}
    \Psi(0) & = z^{2} \\
    tH\Psi(0) & = t\left(2 x^{3}-4 x^{2}+2 x \right)\\
    \frac{t^{2}}{2!} H^{2}\Psi(0) & = \frac{t^{2}}{2!}\left(6 x^{4}-26 x^{3}+30 x^{2}-10 x \right)\\
    \frac{t^{3}}{3!}H^{3}\Psi(0) & =\frac{t^{3}}{3!}\left( 24 x^{5}-174 x^{4}+366 x^{3}-286 x^{2}+70 x \right) \\
    \frac{t^{4}}{4!}H^{4}\Psi(0) &=\frac{t^{4}}{4!} \left( 120 x^{6}-1296 x^{5}+4362 x^{4}-5954 x^{3}+3410 x^{2}-642 x \right)
    \end{align}
    [/tex]
    (Why doesn't the tex work?)
    The fourth order expansion of the exponential becomes
    [tex]
    \begin{align}
    \Psi(t)_{(4)} =& 5t^{4}x^{6}+(-54t^{4}+4t^{3})x^{5}+(-29t^{3}+3t^{2}+(727/4)t^{4})x^{4}+(2t+61t^{3}-(2977/12)t^{4}-13t^{2})x^{3} \notag \\
    &+(-4t+1+15t^{2}+(1705/12)t^{4}-(143/3)t^{3})x^{2}+((35/3)t^{3}-(107/4)t^{4}+2t-5t^{2})x
    \end{align}
    [/tex]
    (the first 4 is to show that it is from the fourth order expansion). A quick calculation show that ## \sum_{n=1}^{4}\psi_{(4),n}(t) = 1 ## however ## \sum_{n=1}^{4}|\psi_{(4),n}(t)| \neq 1##.
    Looking at the solution of $\Psi$ to different orders I find
    [tex]
    \begin{align}
    \psi_{(1),2} & < 0 \quad \text{for} \quad t>\frac{1}{4} \\
    \psi_{(2),3} & < 0 \quad \text{for} \quad t>\frac{2}{13} \\
    \psi_{(3),4} & < 0 \quad \text{for} \quad t>\frac{3}{29} \\
    \psi_{(4),5} & < 0 \quad \text{for} \quad t>\frac{2}{27}
    \end{align}
    [/tex]
    That is the probability becomes negative at shorter and shorter time the higher I expand! This seems horrible.
    Is there any reasonably comments on this? I do not know what to do with this :(.


    Solution attempt 2

    Rescaling the coefficients in the starting ordinary differential equation I find
    $$
    \frac{dP}{dt} = -\alpha P - 2\beta P^{2}
    $$
    Which gives the hamiltonian
    $$
    H = \alpha \left(a -a^{\dagger}a\right) + \beta \left(a^{2} - a^{\dagger2}a^{2}\right)
    $$
    Perhaps it is possible to use something like the Lie product formula
    $$
    e^{tH}\Psi = \lim_{N \rightarrow \infty} \left(e^{t \alpha a/N}e^{-t \alpha a^{\dagger}a/N} e^{t\beta a^{2}/N} e^{-t\beta a^{\dagger2}a^{2}/N} \Psi^{1/N}\right)^{N}
    $$
    Defining ## \Psi = x^{n} ## using some known identities for the generators of scaling and translation and some calculations I find
    [tex]
    \begin{align}
    e^{-t \alpha a^{\dagger}a/N } \Psi^{1/N} &= \Psi^{1/N} \exp\left( - t\alpha \frac{n}{N^{2}} \right) = \Psi^{1/N}A_{\alpha} \\
    e^{-t\beta a^{\dagger2}a^{2}/N} \Psi^{1/N} &= \Psi^{1/N}\exp\left( - t\beta \frac{n/N (n/N-1)}{N} \right) = \Psi^{1/N}A_{\beta} \\
    e^{t \alpha a/N}\Psi^{1/N} &= \left(x + t\alpha/N\right)^{n/N} \\
    e^{t\beta a^{2}/N} \left(x + t\alpha/N\right)^{n/N} & =\left(x + t\alpha/N\right)^{n/N} \sum_{k=0}^{\infty}\frac{(n/N)^{\underbar{2k}}}{\left(Nx+t\alpha\right)^{2k}k!}\left(tN\beta \right)^{k}
    \end{align}
    [/tex]
    (is there a limit on how many tex I can use?)
    This is how far I've gotten with this approach. I don't even know if it is okay to use this Lie product formula.
    Any sort of help would be very much appriciated :). Even vague ideas are welcome.
     
    Last edited by a moderator: Sep 16, 2014
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Ordinarty differential equations
  1. Differential equation (Replies: 2)

  2. Differential equation (Replies: 3)

  3. Differential equation (Replies: 15)

Loading...