# 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.

## 3 Answers

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