Hello Veeky Forums. I am going to run through an argument in quantum mechanics and I want to see if you can help me say more or see if I have made any poor assumptions along the line.
The topic is computing absorption probabilities in a quantum mechanical system. Let us first start with an underlying space [math]\Omega [/math] such that state vectors [math]\psi[/math] are elements of [math]L^2(\Omega )[/math]. Let [math]U:L^2(\Omega )\rightarrow L^2(\Omega )[/math] be a unitary transformation governing the evolution of the system after a single time step. Let [math]S\subset\Omega [/math] and [math]S'\subset S[/math] where elements in [math]S[/math] are absorbing boundaries. The goal of this line of thinking is to compute the probability that we eventually observe the particle in [math]S'[/math] and not in [math]S\cap S'^c[/math].
Now let [math]P:L^2(\Omega )\rightarrow L^2(\Omega )[/math] be a projection onto [math]S[/math] and likewise for [math]P'[/math]. These operators are necessarily self-adjoint. Recall the probabilistic interpretation of quantum mechanics which states that if we measure a state [math]\psi[/math] over a set [math]S[/math], the unnormalized state [math]\psi[/math] becomes [math]P\psi [/math] with probability [math]\lVert P\psi\rVert [/math] (the particle is observed) or [math](I-P)\psi [/math] with probability [math]\lVert (I-P)\psi\rVert [/math] otherwise.
For the remainder of these posts I am going to assume [math]\Omega =\{ 1,...,n\}[/math] for ease of notation so that everything may be represented by matrices, although we could certainly extend the coming arguments for more general sets.
Luke Williams
So are you actually going to post something or not? I don't know much about quantum physics but if you want to do some functional analysis I'm in (or will be tomorrow; as a Europoor I'm off to bed now).
Adrian Butler
thanks boss I will keep posting because of you
For convenience, let [math]S'=\{\omega\}[/math] be a single element set. We will treat the subset [math]S[/math] as an absorbing boundary; that is, we will take a measurement at each element of [math]S[/math]. If we find the particle there, then we will terminate the experiment. Otherwise, we will continue.
Josiah Bennett
I'm not sure what you mean by "absorbing boundary". From what you say, it seems you want to describe properties of [math] M_{\omega_k} \left( U \left( M_{\omega_{k-1}} ... M_{\omega_1} \left( \psi \right) \right) \right), [/math] where [math] M [/math] stands for "taking a measurement" and has an uncertain outcome. I'm assuming time passes between measurements as you wouldn't have introduced [math] U [/math] otherwise.
Owen Morgan
Think about the absorbing boundary in terms of a random walk. If you end up at the absorbing boundary, you terminate the process, otherwise you keep walking. This is made a little more complicated in the quantum domain since you need to measure the state to see whether the particle will be absorbed or not. In between the measurements we evolve the state one step by [math]U[/math], I probably should have made that clear.
Recall we want to find the probability that the particle is EVENTUALLY absorbed in [math]S'[/math] but not in [math]S[/math]. If the particle is absorbed, then it will be absorbed at some time [math]t[/math] and no other. Thus, we can describe the total absorption probability [math]\mathcal{P}[/math] as:
Here, [math]\psi _0[/math] is some initial condition, [math]\psi _1[/math] is a unit vector such that [math]\psi _1\psi _1^T=P'[/math]. To be clear, [math]v^T[/math] is the conjugate transpose of [math]v[/math].
Cameron Anderson
I just want to explain the previous formula briefly. Basically the eventual probability of absorption can be time. The process can be thought of as a two step process: 1) Evolve the state by [math]U[/math] 2) Measure the state by [math]P[/math] Each summand in [math]\mathcal{P}[/math] can be thought of as the amplitude at [math]S'[/math] at time [math]t[/math] before absorption.
One way to compute this is to construct generating functions. We consider the generating function [math]f(z)[/math] defined as follows:
It may not immediately be obvious how to get from [math]f(z)[/math] to [math]\mathcal{P}[/math] but don't worry, that will be addressed in the next post. We can abandon bra-ket notation in the generating function and write this out entirely in matrix/vector notation:
We can pull terms out to focus on a matrix valued generating function [math]F(z)[/math] satisfying [math]f(z)=\psi _1^TUF(z)\psi _0[/math].
Nolan Thompson
If we stop the experiment when we hit the absorbing boundary, shouldn't we use [math] I-P [/math] in the equation for [math] \mathcal{P} [/math]? Other than that I have no trouble understanding the equation.
>Basically the eventual probability of absorption can be time. I don't know what you mean by this, however. Did you forget a word or two? I'm assuming you're talking about how we want to know the probability of eventual absorption, meaning we have to consider all times [math] t [/math].
And it looks like [math] \mathcal{P} [/math] is just [math] f(1) \cdot \bar{f}(1)[/math].
Ryder Ross
Ignore the bit about [math] f(1) \cdot \bar{f}(z) [/math]. I'm retarded. Go on.
Jaxon Campbell
yes yes you're right that using my current notation I should be using [math]I-P[/math] and not [math]P[/math]. However, to simplify notation in the long run, let us redefine [math]P[/math] to project OFF of [math]S[/math] such that my formula is right. However, this means that we should now write [math]\psi _1\psi _1^T=I-P'[/math] where [math]P'[/math] projects OFF of [math]S'[/math].
Let us look more carefully at the function [math]F(z)[/math]. This function can be written as the following series:
Clearly, the eigenvalues of [math]PU[/math] have absolute value less than or equal to 1. However, I will make the assumption that [math]PU[/math] has no eigenvalues with absolute value 1. In practice this is a reasonable assumption to make and it covers a wide variety of cases. In practice, this means that any initial state is totally absorbed by [math]S[/math] in the limit although if [math]S'\ne S[/math] we will have [math]\mathcal{P}\le 1[/math].
The first utility of this assumption is to write this matrix generating function as [math]F(z)=(I-PUz)^{-1}[/math]. To see this, just consider the Taylor expansion and and treat it like a geometric series. By our assumption, this is a well defined function and analytic for [math]|z|\le 1[/math].
