Kets and the Hilbert space
Contents:
# The Hilbert space
Looking at that $i$ in $|S_y;\pm\rangle$ from the Stern-Gerlach notes, the Hilbert space must be a complex vector space.
# Kets
- A physical state is represented by a state vector $\to$ called a “ket”: $|\alpha\rangle$
- It contains complete info about the state.
We can add two kets:
$$ |\alpha\rangle + |\beta\rangle = |\gamma\rangle $$
or constant multiply:
$$ c|\alpha\rangle = |\alpha\rangle c, \qquad c \in \mathbb{C} $$
Physics says $|\alpha\rangle$ and $c|\alpha\rangle$ ($c \neq 0$) represent the same physical state $\implies$ only the “direction” in this vector space is of any significance.
# Observables and eigenkets
An observable is an operator $\to \hat A$. The operator acts on a ket from the left.
$$ \hat A\cdot\left(|\alpha\rangle\right) = \hat A|\alpha\rangle $$
In some special cases, for some “eigenkets”, we have
$$ \hat A|a'\rangle = a'|a'\rangle, \qquad \hat A|a''\rangle = a''|a''\rangle $$
where $|a'\rangle, |a''\rangle$ are all eigenkets, and $a', a''$ are just numbers $\in \mathbb{C}$ (the eigenvalue of the ket).
The physical state corresponding to an eigenket is called an eigenstate.
Eg.
$$ \begin{aligned} \hat S_z|S_z;+\rangle &= \frac{\hbar}{2}|S_z;+\rangle \\ \hat S_z|S_z;-\rangle &= -\frac{\hbar}{2}|S_z;-\rangle \end{aligned} $$
here, $|S_z;\pm\rangle$ are eigenkets of the operator $\hat S_z$. And,
$$ \hat S_x|S_x;\pm\rangle = \pm\frac{\hbar}{2}|S_x;\pm\rangle $$
↳ same eigenvalues, different states.
# Dimensionality
The dimensionality of the vector space is determined by the no. of alternatives in SG type experiments $\implies$ experiments to measure all measurables. How many alternatives? (For the spin in the SG experiment, two: $S_z = \pm\hbar/2$.)
We are concerned with an $N$-dim. vector space spanned by $N$ eigenkets of observable $\hat A$. Then any arbitrary ket $|\alpha\rangle$ can be written as:
$$ |\alpha\rangle = \sum_{a'} c_{a'}|a'\rangle $$
with $a', a'', \dots, a^{(N)}$ and $c_{a'}$ is a $\mathbb{C}$ number.
Is this unique? Will see.
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