We have state defined at each time \(t\).
\(\Psi (t)\).
We have state \(\Psi (t)\).
\(\psi(x,t)=\langle x\rvert \Psi (t) \rangle\)
This is the wave function.
With discrete time we can use a canonical operator for moving between discrete states in single jumps.
With discrete time there must a countable number of states.
We can index time to the integers.
At time \(0\) we have \(v\)
At time \(1\) we have \(\Psi v\)
At time \(2\) we have \(\Psi \Psi v\)
We can write this as \(\Psi (t_1,t_0)=\Psi ^{t_1-t_0}\)
Time is a linear operator
Instead, we describe the time operator as a Lie group, using Lie algebra.
\(\Psi (t_b-t_a)=e^{(t_b-t_a)X}\)
We can remove a degree of freedom by using norm of 1 for vectors
For each dynamic system we define a set of possible states.
We can describe a state \(v\in V\).
We can describe a system like heads or tails.
This can describe continous position, or an angle.
Sloan’s theorem
We use \(X=iH\), what are the implications of this compared to other choices?
Lie algebras with \(n\times x\)
This loops back? multiple dimensions, infinite, so maybe not?
With continuous time we do not have a single operator to describe movements. There is always one smaller.
With continous time there must be either a single state, or an uncountably infinite number of states.
\(U=M_n^n\)
\(U=(I+\dfrac{1}{n}G_n)^n\)
\(U=\lim_{n\rightarrow \infty }(I+\dfrac{1}{n}G)^n\)
Now:
\(UU^*=I\)
\((I+\dfrac{1}{n}G)(I+\dfrac{1}{n}G)^*=I\)
\((I+\dfrac{1}{n}G)(I+\dfrac{1}{n}G^*)=I\)
\(G=-G^*\)
\(G=iH\)
\(iH=-(iH)^*\)
\(H=H^*\)
\(H\) is Hermitian
\(U=\lim_{n\rightarrow \infty }(I+\dfrac{1}{n}iH)^n\)
This isn’t quite right, need defined for different time jumps.
Why? What’s the interpretation here? Is this an assumption, or just a modelling choice?
\(\Psi (t_b-t_a)^* \Psi (t_b-t_a)=e^{(t_b-t_a)X^*} e^{(t_b-t_a)X}\)
\(\Psi (t_b-t_a)^* \Psi (t_b-t_a)=e^{(t_b-t_a)(X^*+X)}\)
\(X=iH\)
\(\Psi (t_b-t_a)^* \Psi (t_b-t_a)=e^{(t_b-t_a)(-iH+iH)}=I\)
\(\Psi (t_b-t_a)=e^{(t_b-t_a)iH}\)
\(v(t_b)=e^{(t_b-t_a)X}v(t_a)\)
\(v(t+\delta )=e^{\delta X}v(t)\)
\(v(t+\delta )=(I+\delta X)v(t)\)
\(\dfrac{v(t+\delta )-v(t)}{\delta }=Xv(t)\)
\(\dfrac{\delta v(t)}{\delta t}=Xv(t)\)
\(\dfrac{\delta v(t)}{\delta t}=iHv(t)\)
\(E=ih\dfrac{\delta }{\delta t}\)
\(Ev(t)=Hv(t)\)
We can add Plank’s constant, due to the arbitrary scaling of time.
Result of spin-statistics theorem?