Subject: Time Evolution Operator

Post date: April 22, 2017

In this page we will review the basic concepts of propagation in the context of

quantum mechanics. Beginning with the Schrodinger equation:

\begin{eqnarray}

i\partial_t U(t,t_0) |\psi(0)\rangle = H(t)U(t,t_0)|\psi(t_0)\rangle, \,\,\,\, \forall |\psi(t_0)\rangle

\end{eqnarray}

where \(U\) is the unitary time-evolution operator, \(|\psi\rangle\) the wave function, and \(H\) is the Hamiltonian,

we identify that

\begin{eqnarray}

i\partial_t U(t,t_0) = H(t)U(t,t_0)

\end{eqnarray}

should hold as an identity, with the initial condition of \(U(t_0,t_0)= 1\).

In reality the system has a time-independent Hamiltonian part

and likely a perturbed component that varies in time, the two constitute \(H(t)\) (\)=H_0 V(t)\)).

Thus,

\begin{eqnarray}

U(t+\delta,t_0) - U(t,t_0) &=& - i \delta H(t) U(t,t_0) \nonumber \\

U(t+\delta,t_0) &=& (1 - i \delta H(t)) U(t,t_0),

\end{eqnarray}

iterating the above identity with discretized time interval \([t_0,t]\),

\begin{eqnarray}

U(t,t_0) &=& \prod_{m=1}^{m=N} (1 - i \delta_m H(t_m)) U(t_0,t_0) \nonumber \\

U(t_0+ N\delta,t_0) &=& \prod_{m=1}^{m=N} (1 - i \delta H(m\delta)) \nonumber \\

&=& 1 -i \delta \sum_m H(m\delta)) - \delta^2 \sum_{n>m} H(n\delta) H(m\delta) \nonumber \\

&+& \dots

\end{eqnarray}

Since \(\sum_{n>m} + \sum_{n<m} \rightarrow \sum_{n,m}\) in the limit of \(\delta \rightarrow 0\) we have

\begin{eqnarray}

U(t_0 + N\delta,t_0) &=& 1 -i \delta \sum_m H(m\delta)

- \frac{1}{2} \delta^2 \sum_{n,m} \hat{T}[H(n\delta) H(m\delta)] \nonumber \\

&+& \dots \nonumber \\

&\rightarrow& 1 -i \int_{t_0}^t ds H(s) \nonumber \\

&+& \frac{(-i)^2}{2!} \int_{t_0}^t \int_{t_0}^t ds_1 ds_2 \hat{T} [H(s_1) H(s_2)] \nonumber \\

&+& \dots \nonumber \\

U(t ,t_0) &:=& \hat{T} e^{-i\int_{t_0}^t H(s) ds}

\end{eqnarray}

where \(\hat{T}\) stands for the time ordering operator and means the operators must be written in order of their times with earlier time on the right. The last line is a symbolic representation. The exponential $$e^{i(\dots)}$$ with \(\dots\) a Hermitian operator (that is "real" eigenvalues),

emphasizes that the operator \(U\) is a unitary operator (preserving the norm).

However, as one can visually see, the truncated expansion does not seem to be

of norm one. This result in the norm of the state

\(U(t,t_0)|\psi\rangle\) to vary from one. Thus the expansion must be accompanied by a denominator \(\frac{U|\psi\rangle}{\langle \psi | U^\dagger U |\psi\rangle}\).

In the adiabatic regime (non-degenerate), Gell-Mann and Low proved that the state

\(\frac{U|\psi\rangle}{\langle \psi | U^\dagger U |\psi\rangle}\)is an eigen state of the \(H(t)\) as the

time-varying part of the Hamiltonian is adiabatically

turned on . This is in particular important in the study of the ground state

of interacting many-body systems where the \(|\psi\rangle\) is the ground state of the non-interacting system and is

adiabatically perturbed to an interacting regime.

Subject: Bias-Variance Trade-Off

Post date: May 14, 2017

Here is a quick demonstration of Bias-Variance Trade-Off.

I am going to use some actual data. The goal is to fit a polynomial function to the data points.

I call the target data as \(y\). The index to the data \(x\) is normalized so they run from \([0,1]\).

We are going to first pick a subset, the training set, and use them to find the correct

parameters of the model. The model is a polynomial function, \(f(x)=a_0 + \dots a_N x^N\).

The maximum power \(N\) that needs to be used in the model is unknown. So, we are going to

use the Bias-Variance Trade-Off concept to find what is the best power to be used.

Here I use half of the data points to train the model. The data looks something like this:

Using this set I find the parametrs $A$ such that $$X(x) \,\, A= Y_t.$$ here \(X\) is a matrix width rows that goes like \(\left[x^0 \dots \, x^{N} \right] \) points according to the maximum power of the polynomial. For a max power of \(4\) the plot looks like

repeating the process for a set of models created as explained above we can find that the R-MSE (Root of the Mean Square Error) looks something like this: