# Quantum mechanics

## State evolution

### Indexing states to time

We have state defined at each time $$t$$.

$$\Psi (t)$$.

### Wave functions

We have state $$\Psi (t)$$.

$$\psi(x,t)=\langle x\rvert \Psi (t) \rangle$$

This is the wave function.

### Schrodinger

#### Discrete time

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 $$\Psi (t_1,t_0)=\Psi ^{t_1-t_0}$$

### Representation theory for the time group

Time is a linear operator

Instead, we describe the time operator as a Lie group, using Lie algebra.

$$\Psi (t_b-t_a)=e^{(t_b-t_a)X}$$

#### States are vectors

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 $$v\in V$$.

#### Finite state spaces

We can describe a system like heads or tails.

#### Infinite state spaces

This can describe continous position, or an angle.

Sloan’s theorem

### Continous time with Lie algebra

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.

### Unitary time

Why? What’s the interpretation here? Is this an assumption, or just a modelling choice?

$$\Psi (t_b-t_a)^* \Psi (t_b-t_a)=e^{(t_b-t_a)X^*} e^{(t_b-t_a)X}$$

$$\Psi (t_b-t_a)^* \Psi (t_b-t_a)=e^{(t_b-t_a)(X^*+X)}$$

$$X=iH$$

$$\Psi (t_b-t_a)^* \Psi (t_b-t_a)=e^{(t_b-t_a)(-iH+iH)}=I$$

$$\Psi (t_b-t_a)=e^{(t_b-t_a)iH}$$

### The time-depedendent general Schrödinger equation

$$v(t_b)=e^{(t_b-t_a)X}v(t_a)$$

$$v(t+\delta )=e^{\delta X}v(t)$$

$$v(t+\delta )=(I+\delta X)v(t)$$

$$\dfrac{v(t+\delta )-v(t)}{\delta }=Xv(t)$$

$$\dfrac{\delta v(t)}{\delta t}=Xv(t)$$

$$\dfrac{\delta v(t)}{\delta t}=iHv(t)$$

### The energy operator and the time-indepedendent general Schrödinger equation

$$E=ih\dfrac{\delta }{\delta t}$$

$$Ev(t)=Hv(t)$$

## Other

### Plank’s constant

We can add Plank’s constant, due to the arbitrary scaling of time.

### Heisenberg’s uncertainty principle

Result of spin-statistics theorem?