Review of Linear Algebra in Quantum Mechanics (I)

This series of posts is just a refresh of linear algebra basics in the language of quantum mechanics. One of the motivations is that in many books like Prof. Georgi's "Lie Algebra in Particle Physics", linear algebra is directly presented in quantum mechanics notations.

Vector Space

The fundamental concept in linear algebra is the so-called vector space (or linear space). It is a set of elements called vectors that can be added together or multiplied by scalars in the field $\mathbb{F}$. When the scalars are complex numbers, we can denote the vectors by kets $|\,\rangle$. So in the language of quantum mechanics, the basic definition of vector space over $\mathbb{C}$ can be formulated as follows:

1. $|\psi\rangle + |\phi\rangle = |\phi\rangle + |\psi\rangle$.
2. $\Big(|\psi\rangle + |\phi\rangle\Big)+|\chi\rangle=|\psi\rangle+\Big(|\phi\rangle + |\chi\rangle\Big)$.
3. There exists an $\mathbf{0}$ such that $|\psi\rangle + \mathbf{0}=|\psi\rangle$ for every $|\psi\rangle$.
4. For every $|\psi\rangle$, there is $-|\psi\rangle$ such that $|\psi\rangle+\Big(-|\psi\rangle\Big)=\mathbf{0}$.
5. $c_1\Big(c_2 |\psi\rangle\Big)=\Big(c_1c_2\Big)|\psi\rangle$.
6. $1\,|\psi\rangle = |\psi\rangle$. 
7. $c\Big(|\psi\rangle+|\phi\rangle\Big)=c|\psi\rangle + c|\phi\rangle$.
8. $\Big(c_1+c_2\Big)|\psi\rangle = c_1|\psi\rangle + c_2|\psi\rangle$.

Note that we denote the zero vector in the 3rd axiom by $\mathbf{0}$ instead of $|0\rangle$. This is because $|0\rangle$ usually represents the vaccum state in quantum mechanics. For an annihilation operator $a$, we have $a|0\rangle=\mathbf{0}$.

To better understand the eight axioms, let's do some exercises.

[Exercise 1] Prove: $\mathbf{0}$ is unique.

[Solution] Suppose both $\mathbf{0}$ and $\mathbf{0'}$ are zero vectors, then \begin{eqnarray}\mathbf{0} &=& \mathbf{0} + \mathbf{0'}\quad\quad\text{(3rd axiom)}\\&=& \mathbf{0'} + \mathbf{0}\quad\quad\text{(1st axiom)}\\ &=& \mathbf{0'}\,.\quad\quad\quad\,\text{(3rd axiom)}\end{eqnarray}

[Exercise 2] Prove: $-|\psi\rangle$ is unique.

[Solution] Suppose $|\psi\rangle + |\phi\rangle =\mathbf{0}$ and $|\psi\rangle + |\chi\rangle =\mathbf{0}$, then \begin{eqnarray}|\phi\rangle &=& |\phi\rangle + \mathbf{0}\quad\quad\quad\quad\quad\quad\quad\quad\text{(3rd axiom)}\\&=&  |\phi\rangle + \Big(|\psi\rangle + |\chi\rangle \Big)\\ &=& \Big(|\phi\rangle + |\psi\rangle\Big) + |\chi\rangle\quad\quad\quad\,\,\,\text{(2nd axiom)}\\&=&|\chi\rangle + \Big(|\psi\rangle+|\phi\rangle\Big)  \quad\quad\quad\,\,\,\text{(1st axiom)} \\&=&|\chi\rangle + \mathbf{0} \\ &=& |\chi\rangle\,.\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\text{(3rd axiom)} \end{eqnarray}

[Exercise 3] Prove: $0\,|\psi\rangle = \mathbf{0}$.

[Solution] \begin{eqnarray} 0\,|\psi\rangle &=& 0\,|\psi\rangle + \mathbf{0}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\text{(3rd axiom)}\\&=&0\,|\psi\rangle +\bigg(0\,|\psi\rangle + \Big(-0|\psi\rangle\Big)\bigg)\quad\quad\,\text{(4th axiom)}\\&=&\bigg(0\,|\psi\rangle +0\,|\psi\rangle\bigg) + \Big(-0|\psi\rangle\Big)\quad\quad\,\text{(2nd axiom)}\\&=&0\,|\psi\rangle+\Big(-0|\psi\rangle\Big)\quad\quad\quad\quad\quad\quad\quad\text{(8th axiom)}\\&=&\mathbf{0}\,.\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\text{(4th axiom)} \end{eqnarray}

[Exercise 4] Prove: $(-1)\,|\psi\rangle = -|\psi\rangle$.

[Solution] \begin{eqnarray} \mathbf{0} &=& 0\,|\psi\rangle \quad\quad\quad\quad\quad\quad\quad\quad\,\text{(Exercise 3)}\\ &=& (1+(-1))\,|\psi\rangle  \\ &=& 1\,|\psi\rangle + (-1)\,|\psi\rangle \quad \quad \quad\,\,\text{(8th axiom)}\\ &=& |\psi\rangle + (-1)\,|\psi\rangle\,,\quad \quad\quad\,\,\text{(6th axiom)}\end{eqnarray} which suggests that $(-1)\,|\psi\rangle = -|\psi\rangle$ by the 4th axiom and Exercise 2.

[Exercise 5] Prove: $c\,\mathbf{0} = \mathbf{0}$.

[Solution] \begin{eqnarray} c\,\mathbf{0} &=& c\,\bigg(|\psi\rangle + \Big(-|\psi\rangle\Big)\bigg) \quad\quad\quad\text{(4th axiom)}\\ &=& c\,|\psi\rangle + c \Big(-|\psi\rangle\Big)\quad\quad\quad\quad \text{(7th axiom)} \\ &=& c\,|\psi\rangle + c(-1)\,|\psi\rangle \quad \quad \quad\quad\,\text{(Exercise 4)}\\ &=& c\,|\psi\rangle + (-c)\,|\psi\rangle \quad \quad\quad\quad\,\,\,\,\text{(5th axiom)}\\&=& 0\,|\psi\rangle \quad \quad\quad\quad\quad \quad\quad\quad\quad\,\,\text{(8th axiom)}\\&=&\mathbf{0}\,.\quad \quad\quad\quad\quad \quad\quad\quad\quad\quad\,\text{(Exercise 3)}\end{eqnarray}

Note:

• To prove the properties as in Exercise 1-5 above, we do have used all the eight axioms. 
• There is a claim that the first axiom of commutativity can be derived from the remaining seven axioms. For example, one can find such proof in this post. But there is a flaw in the proof: $|\psi\rangle + |\phi\rangle - \Big(|\phi\rangle + |\psi\rangle\Big)=\mathbf{0}$ does not lead to $|\psi\rangle + |\phi\rangle = |\phi\rangle + |\psi\rangle$. It requires an assumption that $-|\psi\rangle +|\psi\rangle=0$, which is equivalent to the first axiom as compared with the fourth axiom.