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I'm studying the Bloch Sphere and just wanted to ask what this notation means: $|\psi\rangle = \alpha|1\rangle$ for example

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I'm just not familiar with the notation in this context if anyone could explain it to me or point me to the right direction I'd appreciate it.

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This is the Dirac notation. In Quantum Mechanics states of a system are unit rays on a Hilbert space - a special kind of vector space with inner product. In turn, they can be characterized by unit vectors in this vector space. It is usual to denote such a state vector as something like $|\psi\rangle$.

If $|\chi\rangle$ is another state vector, the inner product in the Hilbert space is denoted $\langle \chi |\psi\rangle$.

In turn, we can have bases. An orthonormal basis here is a set of vectors $|\phi_n\rangle$ such that $$\langle \phi_n|\phi_m\rangle=\delta_{nm}.$$

The basis must be complete in the sense that if $|\chi\rangle$ is orthogonal to all $|\phi_n\rangle$, i.e., $\langle \phi_n | \chi\rangle=0$ then $|\chi\rangle=0$.

Now, when we need an infinite set of such $|\phi_n\rangle$ to form a complete set, the Hilbert space is infinite dimensional. When we have a complete set with a finite number $D$ of vectors, the Hilbert space has dimension $D$.

In particular, we can have a $2$-dimensional Hilbert space with a basis $\{|0\rangle,|1\rangle\}$. This is just notation, we could have written $\{|\phi_1\rangle,|\phi_2\rangle\}$. One concrete example of this in nature would be the description of the spin degrees of freedom of a non-relativistic spin $1/2$ particle like the electron.

Now one arbitrary state $|\psi\rangle$ can be written as

$$|\psi\rangle = a |0\rangle + b|1\rangle.$$

Furthermore, we normalize states as I already said, so that we require $\langle \psi|\psi\rangle = 1$, this means that upon using orthonormality of the basis:

$$|a|^2+|b|^2=1.$$

This in turn means that $a = e^{i\phi_1}\cos\theta$ and $b=e^{i\phi_2}\sin\theta$ for some $\phi_1,\phi_2,\theta$. We can choose $\phi_1=0$ though. To prove this, notice that since $|\psi\rangle$ must have unit norm, we can multiply it by any phase $e^{i\phi}$ and represent the same physical state. Thus we exchange $|\psi\rangle$ by $e^{-i\phi_1}|\psi\rangle$. This makes the state become

$$|\psi\rangle = \cos\theta |0\rangle + e^{i\phi}\sin\theta|1\rangle.$$

That is the reason the state can always be written like this.

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This is known as Dirac notation, also known as bra-ket notation. There's a good explanation on Wikipedia, and it is discussed in depth in all QM textbooks - look for either of those names in the index to find it.

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  • $\begingroup$ @EmilioPisanty ... and you were right again. $\endgroup$ – ZeroTheHero Sep 21 '18 at 14:35
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You need both angles so that you can generate, for instance, states along $\hat y$: $$ \vert +\hat y\rangle =\frac{1}{\sqrt{2}} \left(\vert +\rangle_z -i \vert -\rangle_z\right) $$ which corresponds in your notation to $\theta=\pi/4$ and $\phi=-\pi/2$.

Please note that the standard parametrization is usually in terms of the half-angle $\theta$, i.e. $$ \vert\psi\rangle=\cos\frac{\theta}{2}\vert 0\rangle + e^{i\phi}\sin\frac{\theta}{2}\vert 1\rangle $$ where $\vert 0\rangle=\vert +\rangle_z$ and $\vert 1\rangle=\vert -\rangle_z$ and $\vert \pm\rangle_z$ the eigenstates of $\sigma_z$.

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  • $\begingroup$ It seems we continue to read questions very differently ;-). $\endgroup$ – Emilio Pisanty Sep 21 '18 at 14:31
  • $\begingroup$ @EmilioPisanty no big deal. I'm quite tired so you're probably right again. $\endgroup$ – ZeroTheHero Sep 21 '18 at 14:32
  • $\begingroup$ Thanks, guys much appreciated. I'd upvote you guys but I don't have a reputation built up yet :) $\endgroup$ – M00N KNIGHT Sep 21 '18 at 14:34

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