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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


    for some \( |\phi\rangle, |\psi\rangle \in\) Screen Shot 2021-11-22 at 3.40.11 AM.png. 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

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

    and suppose that

    2 & 0 \\
    0 & 3

    We calculate

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

    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

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

    which leads to

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

    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_{1} & 0 & \cdots & 0 \\
    0 & a_{2} & \cdots & 0 \\
    \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & \cdots & a_{N}

    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


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


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


    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 \(A B=B A\).

    &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|


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

    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.

    This page titled 1.2: Operators in Hilbert Space is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pieter Kok via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.