# 1.2: Operators in Hilbert Space

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The objects $$|\psi\rangle$$ are vectors in a Hilbert space. We can imagine applying rotations of the vectors, rescaling, permutations of vectors in a basis, and so on. These are described mathematically as operators, and we denote them by capital letters A, B, C, etc. In general we write

$A|\phi\rangle=|\psi\rangle\tag{1.6}$

for some $$|\phi\rangle, |\psi\rangle \in$$ . It is important to remember that operators act on all the vectors in Hilbert space. Let $$\left\{\left|\phi_{j}\right\rangle\right\}_{j}$$ be an orthonormal basis. We can calculate the inner product between the vectors $$\left|\phi_{j}\right\rangle$$ and A $$\left|\phi_{k}\right\rangle$$:

$\left\langle\phi_{j}\right|\left(A\left|\phi_{k}\right\rangle\right)=\left\langle\phi_{j}|A| \phi_{k}\right\rangle \equiv A_{j k}\tag{1.7}$

The two indices indicate that operators are matrices.

As an example, consider two vectors, written as two-dimensional column vectors

$\left|\phi_{1}\right\rangle=\left(\begin{array}{l} 1 \\ 0 \end{array}\right), \quad\left|\phi_{2}\right\rangle=\left(\begin{array}{l} 0 \\ 1 \end{array}\right)\tag{1.8}$

and suppose that

$A=\left(\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right)\tag{1.9}$

We calculate

$A_{11}=\left\langle\phi_{1}|A| \phi_{1}\right\rangle=(1,0) \cdot\left(\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right)\left(\begin{array}{l} 1 \\ 0 \end{array}\right)=(1,0) \cdot\left(\begin{array}{l} 2 \\ 0 \end{array}\right)=2\tag{1.10}$

Similarly, we can calculate that $$A_{22}=3$$, and $$A_{12}=A_{21}=0$$ (check this). We therefore have that $$A\left|\phi_{1}\right\rangle=2\left|\phi_{1}\right\rangle$$ and $$A\left|\phi_{2}\right\rangle=3\left|\phi_{2}\right\rangle$$.

Complex numbers $$a$$ have complex conjugates $$a^{*}$$ and vectors $$|\psi\rangle$$ have dual vectors $$\langle\phi|$$. Is there an equivalent for operators? The answer is yes, and it is called the adjoint, or Hermitian conjugate, and is denoted by a dagger $$\dagger$$. The natural way to define it is according to the rule

$\langle\psi|A| \phi\rangle^{*}=\left\langle\phi\left|A^{\dagger}\right| \psi\right\rangle\tag{1.11}$

for any $$|\phi\rangle$$ and $$|\psi\rangle$$. In matrix notation, and given an orthonormal basis $$\left\{\left|\phi_{j}\right\rangle\right\}_{j}$$, this becomes

$\left\langle\phi_{j}|A| \phi_{k}\right\rangle^{*}=A_{j k}^{*}=\left\langle\phi_{k}\left|A^{\dagger}\right| \phi_{j}\right\rangle=A_{k j}^{\dagger}\tag{1.12}$

So the matrix representation of the adjoint $$A^{\dagger}$$ is the transpose and the complex conjugate of the matrix A, as given by $$\left(A^{\dagger}\right)_{j k}=A_{k j}^{*}$$. The adjoint has the following properties:

1. $$(c A)^{\dagger}=c^{*} A^{\dagger}$$,
2. $$(A B)^{\dagger}=B^{\dagger} A^{\dagger}$$,
3. $$(|\phi\rangle)^{\dagger}=\langle\phi|$$.

Note the order of the operators in 2: AB is generally not the same as BA. The difference between the two is called the commutator, denoted by

$[A, B]=A B-B A\tag{1.13}$

For example, we can choose

$A=\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) \quad \text { and } \quad B=\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\tag{1.14}$

\begin{aligned}{} [A, B] &=\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)-\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right)=\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)-\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right) \\ &=\left(\begin{array}{cc} 0 & -2 \\ 2 & 0 \end{array}\right) \neq 0 . \end{aligned}\tag{1.15}

Many, but not all, operators have an inverse. Let $$A|\phi\rangle=|\psi\rangle$$ and $$B|\psi\rangle=|\phi\rangle$$. Then we have

$B A|\phi\rangle=|\phi\rangle \quad \text { and } \quad A B|\psi\rangle=|\psi\rangle\tag{1.16}$

If Eq. (1.16) holds true for all $$|\phi\rangle$$ and $$|\psi\rangle$$, then $$B$$ is the inverse of $$A$$, and we write $$B=A^{-1}$$. An operator that has an inverse is called invertible. Another important property that an operator may possess is positivity. An operator is positive if

$\langle\phi|A| \phi\rangle \geq 0 \quad \text { for all }|\phi\rangle\tag{1.17}$

We also write this as $$A \geq 0$$.

From the matrix representation of operators you can easily see that the operators themselves form a linear vector space:

1. $$A+B=B+A$$,
2. $$A+(B+C)=(A+B)+C$$,
3. $$a(A+B)=a A+a B$$,
4. $$(a+b) A=a A+b A$$,
5. $$a(b A)=(a b) A$$.

We can also define the operator norm $$\|A\|$$ according to

$\|A\|=\sqrt{\operatorname{Tr}\left(A^{\dagger} A\right)} \equiv \sqrt{\sum_{i j} A_{i j}^{*} A_{j i}},\tag{1.18}$

which means that the linear vector space of operators is again a Hilbert space. The symbol Tr(.) denotes the trace of an operator, and we will return to this special operator property later in this section.

Every operator has a set of vectors for which

$A\left|a_{j}\right\rangle=a_{j}\left|a_{j}\right\rangle, \quad \text { with } a_{j} \in \mathbb{C}\tag{1.19}$

This is called the eigenvalue equation (or eigenequation) for $$A$$, and the vectors $$\left|a_{j}\right\rangle$$ are the eigenvectors. The complex numbers $$a_{j}$$ are eigenvalues. In the basis of eigenvectors, the matrix representation of $$A$$ becomes

$A=\left(\begin{array}{cccc} a_{1} & 0 & \cdots & 0 \\ 0 & a_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{N} \end{array}\right)\tag{1.20}$

When some of the $$a_{j} \mathrm{s}$$ are the same, we speak of degenerate eigenvalues. When there are $$n$$ identical eigenvalues, we have $$n$$-fold degeneracy. The eigenvectors corresponding to this eigenvalue then span an $$n$$-dimensional subspace of the vector space. We will return to subspaces shortly, when we introduce projection operators.

For any orthonormal basis $$\left\{\left|\phi_{j}\right\rangle\right\}_{j}$$ we have

$\left\langle\phi_{j}|A| \phi_{k}\right\rangle=A_{j k}\tag{1.21}$

which can be written in the form

$A=\sum_{j k} A_{j k}\left|\phi_{j}\right\rangle\left\langle\phi_{k}\right|\tag{1.22}$

For the special case where $$\left|\phi_{j}\right\rangle=\left|a_{j}\right\rangle$$ this reduces to

$A=\sum_{j} a_{j}\left|a_{j}\right\rangle\left\langle a_{j}\right|\tag{1.23}$

This is the spectral decomposition of $$A$$. When all $$a_{j}$$ are equal, we have complete degeneracy over the full vector space, and the operator becomes proportional (up to a factor $$a_{j}$$) to the identity $$\mathbb{I}$$.

Note that this is independent of the basis $$\left\{\left|a_{j}\right\rangle\right\}$$. As a consequence, for any orthonormal basis $$\left\{\left|\phi_{j}\right\rangle\right\}$$ we have

$\mathbb{I}=\sum_{j}\left|\phi_{j}\right\rangle\left\langle\phi_{j}\right|\tag{1.24}$

This is the completeness relation, and we will use this many times in our calculations.

## Lemma

If two non-degenerate operators commute $$([A, B]=0)$$, then they have a common set of eigenvectors.

Proof

Let $$A=\sum_{k} a_{k}\left|a_{k}\right\rangle\left\langle a_{k}\right|$$ and $$B=\sum_{j k} B_{j k}\left|a_{j}\right\rangle\left\langle a_{k}\right|$$. We can choose this without loss of generality: we write both operators in the eigenbasis of $$A$$. Furthermore, $$[A, B]=0$$ implies that $$A B=B A$$.

\begin{aligned} &A B=\sum_{k l m} a_{k} B_{l m}|a_{k}\rangle\langle a_{k} \mid a_{l}\rangle\langle a_{m}|=\sum_{l m} a_{l} B_{l m}| a_{l}\rangle\langle a_{m}| \\ &B A=\sum_{k l m} a_{k} B_{l m}|a_{l}\rangle\langle a_{m} \mid a_{k}\rangle\langle a_{k}|=\sum_{l m} a_{m} B_{l m}| a_{l}\rangle\left\langle a_{m}\right| \end{aligned}\tag{1.25}

Therefore

$[A, B]=\sum_{l m}\left(a_{l}-a_{m}\right) B_{l m}\left|a_{l}\right\rangle\left\langle a_{m}\right|=0\tag{1.26}$

If $$a_{l} \neq a_{m}$$ for $$l \neq m$$, then $$B_{l m}=0$$, and $$B_{l m} \propto \delta_{l m}$$. Therefore $$\left\{\left|a_{j}\right\rangle\right\}$$ is an eigenbasis for B.

The proof of the converse is left as an exercise. It turns out that this is also true when $$A$$ and/or $$B$$ are degenerate.

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