# Quantum Mechanics Notation

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

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.

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.

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.

• @EmilioPisanty ... and you were right again. – ZeroTheHero Sep 21 '18 at 14:35

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 $$\sigma_z$$.

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