So far, I have understood that vectors are represented in a basis, and operators are linear maps which map one vector to another in the same space or to a different space.

What I don't understand is this: when we take the inner product of operators with basis vectors it gives components of the operator, so I can multiply the components with the corresponding basis to give a vector — but operators are different from vectors.

What are these "components" of operators that I am getting? Is there a geometrical interpretation for this? $$ A|e_j\rangle = \sum_{i=1}^n A_{ij}|e_i\rangle $$

  • $\begingroup$ Strictly speaking, "operators" are only " linear maps which map one vector to another in the same space," not to other spaces. If the mapping is between two different spaces, then the more general object is a linear "transformation." In other words, a linear transformation $T:V\rightarrow V'$ is a linear operator if and only if $V=V'$. $\endgroup$
    – Buzz
    Commented Sep 11, 2021 at 19:39
  • $\begingroup$ @Buzz In a very dirty way of doing things, for inner-product spaces, I always took an operator to be also defining a mapping from $\mathcal{V}\times\mathcal{V}^\ast$ to $\mathcal{F}$ because given a unique inner-product, a mapping $O: \mathcal{V}\to\mathcal{V}$ naturally induces a map from $\mathcal{V}\times\mathcal{V}^\ast$ to $\mathcal{F}$? $\endgroup$
    – user87745
    Commented Sep 11, 2021 at 20:32

6 Answers 6


Note: I use Einstein summation notation throughout. Any expression which has an index repeated twice should be understood to have an omitted, but implicit summation over that index.


Consider the expression

$$ T|v\rangle $$

where $|v\rangle \in \mathcal{H}$ is an abstract, finite-dimensional vector living in a Hilbert space and $T : \mathcal{H} \rightarrow \mathcal{H}$ is an operator on the Hilbert space which maps old vectors to new vectors. Try to resist the urge to imagine $|v\rangle$ as a column vector and $T$ as a matrix. We have not yet chosen a basis so we would not be able to populate these arrays with numbers yet. That will follow shortly.

Linear algebra teaches us that in a finite dimensional vector space (Hilbert space is a vector space) that we can find an orthonormal set of vectors $\{|e_i\rangle\}$ which are linearly independent and span the vector space. The latter fact means any vector can be written

Expanding a vector into components

$$ |v\rangle = |e_i\rangle {^e[v]_i} $$

Here ${^e[v]_i}$ is the $i^{th}$ component of $v$ with respect to the basis $\{e_i\}$.

Recall that bras like $\langle e_i|$ are dual vectors which act on vectors (kets) and return scalars. Acting a bra on a ket is like taking the dot product of the corresponding vectors, but slightly more generally for non-Euclidean spaces (which we don't deal with in regular quantum mechanics anyways).

We can act the bra $\langle e_j |$ on the left of this expression and see

$$\langle e_j |v \rangle = \langle e_j |e_i \rangle {^e[v]_i} = \delta_{ji}{^e[v]_i} = {^e[v]_j}$$

$$ |v\rangle = |e_i\rangle \langle e_i | v\rangle $$

I've used the orthonormality of the basis for $\langle e_j |e_i \rangle = \delta_{ji}$. So we see that the $j^{th}$ component of the vector $|v\rangle$ with respect to basis $\{e_i\}$ (which we call ${^e[v]_j}$) is exactly $\langle e_j | v \rangle$. In Dirac notation we find components of things by constructing "closed" bracket structures.

Expanding an Operator into Components

How does this work for operators?

$$ T |v \rangle = T |e_j\rangle\langle e_j|v\rangle $$

Here I've just expanded $|v\rangle$ into it's components. Note that it looks like I just "squeezed" the operator $|e_j\rangle \langle e_j| = \sum_j |e_j \rangle \langle e_j|$ in between $T$ and $|v\rangle$. It turns out, because of the completeness of the basis $\{|e_j\rangle\}$, that $|e_j\rangle \langle e_j|$ is equivalent to the identity $I$ so you can squeeze it in anywhere. This is of indescribable utility when it comes to doing linear algebra in Dirac notation.*

Note that the construction $\left(T|v\rangle\right) = |v'\rangle$ is just equivalent to another ket. Let's find the components of this ket by acting on the left with $\langle e_i |$ (Like taking the dot product with $\langle e_i|$).

$$ \langle e_i | T |e_j\rangle \langle e_j |v\rangle $$

Now the quantity $\langle e_i | T |e_j\rangle$ is a scalar and $\langle e_j |v\rangle$ is also a scalar. We call the former quantity the $i, j$ component of the matrix which represents the operator $T$ in the $e$ basis. Suppose $\{|f_i\rangle\}$ is another complete orthonormal basis. We then define

$$ {^e_f [T]_{ij}} = \langle e_i | T | f_j\rangle $$

In case we use the same basis on the left and right we have ${^e_e[T]_{ij}}$ and we might omit the pre-subscript.**

Matrix Expressions

We can then write

$$ \langle e_i | T |e_j\rangle \langle e_j |v\rangle = {^e_e [T]_{ij}} {^e[v]_j} $$

What we have done here is we have taken the original, abstract, and basis-independent expression $T|v\rangle$ and represented it as a more concrete matrix formula with respect to a particular basis.

We can even write it as a true matrix expression (as opposed to an expression on the scalar components):

$$ {^e_e [T]} {^e[v]} $$

Where ${^e_e [T]}$ is an $N\times N$ matrix and ${^e[v]}$ is a length $N$ column vector, both of whose components are described above.

$$ {^e_e [T]} = \begin{pmatrix} {^e_e [T]_{11}} && \ldots && {^e_e [T]_{1N}}\\ \vdots && \ddots && \vdots\\ {^e_e [T]_{N1}} && \ldots && {^e_e [T]_{NN}} \end{pmatrix} = \begin{pmatrix} \langle e_1 |T|e_i\rangle && \ldots && \langle e_1 |T | e_N\rangle\\ \vdots && \ddots && \vdots\\ \langle e_N|T |e_1\rangle && \ldots && \langle e_N |T|e_N\rangle \end{pmatrix} $$

$$ {^e[v]} = \begin{pmatrix} {^e[v]_1} \\ \vdots \\ {^e[v]_N} \end{pmatrix} = \begin{pmatrix} \langle e_1 |v\rangle \\ \vdots \\ \langle e_N |v\rangle \end{pmatrix} $$


Both the abstract and concrete expressions have their usefulness. The abstract expression is useful for theoretical manipulations and explorations, while the concrete expression is better for finding actual numerical solutions to problems. One might argue the coordinate representation is more intuitively clear, but I would argue that the coordinate-free representation provides more flexibility and that it's best if one can build as much intuition for the abstract expressions as for the coordinate expressions.

Direct Answer to the Question

To concretely answer the question: "What is meant by the components of an operator?".

The intuitive answer is that the components of an operator are the matrix elements of the operator when it is represented as a matrix with respect to a particular basis. Mathematically, the components of an operator are the scalar results of squeezing an operator between a bra and a ket. Specifically,

$$ {^e_e [T]_{ij}} = \langle e_i |T|e_j\rangle $$

This is the $i, j$ component of $T$ with respect to basis $\{|e_i\rangle\}$.

A Geometric Interpretation

A geometric interpretation for the components of an operator:

$${^e_e [T]_{ij}} = \langle e_i | T | e_j\rangle$$

We can see that ${^e_e [T]_{ij}}$ gives us the $i^{th}$ component of $|e'_j\rangle = T|e_j\rangle$. That is, how much does $T$ skew the $|e_j\rangle$ basis vector into the $|e_i \rangle$ direction.


*It's also important for using the resolution of the identity that you can take the operator $(|e_i\rangle \langle e_j|$ and "split it up" so that the ket on the left gets "attached" to whatever is to the left of the resolution and the bra on the right gets "attached" to whatever is on the right. To prove this you need to define a few things and prove some things about parentheses. I don't do it here but just want to state that there is a little more under the hood that meets the eye. This fluidity of parentheses (and the ability to not worry about them) is the other thing that makes Dirac notation so great.

**And in fact, most authors never write either of the pre sub or superscripts nor the brackets nor anything. What I write as ${^e_e[T]_{ij}}$ will be seen as $T_{ij}$. While it is usually clear which basis we are working in from context, it can be confusing for people new to the study to understand the difference between $T$ and $T_{ij}$ without the additional notation I have included. Hence this question.


Is there a geometrical interpretation for this?

Yes. Without being too rigorous here, if you identify an operator with a matrix, and a ket with a column vector, then from linear algebra, when you multiply a matrix with a basis column vector (assuming the standard basis), it just picks out the column in the matrix corresponding to the index of the non-zero entry. E.g.

$$ M \hat e_1 = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} a \\ c \\ \end{bmatrix} = \vec\varepsilon_1, $$

$$ M \hat e_2 = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} b \\ d \\ \end{bmatrix} = \vec\varepsilon_2, $$

were $\vec\varepsilon_1$ & $\vec\varepsilon_2$ are the transformed $\hat e_1$ & $\hat e_2$, respectively. In other words, the columns of the matrix are the new basis vectors, expressed in standard basis–components. The action of the matrix can be understood as a reorientation and a rescalling of the coordinate axes. Transforming a vector takes it to its corresponding place in the new basis.

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Note that a vector $\vec v = (x, y)$ can be written as $\vec v = x \hat e_1 + y \hat e_2$, or in matrix notation:

$$ \begin{bmatrix} x \\ y \\ \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} $$

and so

$$ M \begin{bmatrix} x \\ y \\ \end{bmatrix} = xM \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + yM \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = x \vec\varepsilon_1 + y \vec\varepsilon_2 $$

You can do the same for the transformed basis vectors, which in turn recovers the expression you asked about.


$$ A = a_{ij} = a_{(row)(col)} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} $$

then (the $\color{red}{j=1}$ case, iterating over $i$)
$$ A \hat e_{\color{red}1} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} a_{11} \\ a_{21} \\ \end{bmatrix} = a_{1\color{red}1} \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + a_{2\color{red}1} \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = a_{1\color{red}1} \hat e_1 + a_{2\color{red}1} \hat e_2 $$

and (the $\color{red}{j=2}$ case, iterating over $i$)
$$ A \hat e_{\color{red}2} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} a_{12} \\ a_{22} \\ \end{bmatrix} = a_{1\color{red}2} \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + a_{2\color{red}2} \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = a_{1\color{red}2} \hat e_1 + a_{2\color{red}2} \hat e_2 $$

In other words

$$A|e_j\rangle = \sum_{i=1}^n a_{ij}|e_i\rangle$$

is short for

$$ A|e_{\color{red}1}\rangle = a_{1\color{red}1}|e_1\rangle + a_{2\color{red}1}|e_2\rangle \\ A|e_{\color{red}2}\rangle = a_{1\color{red}2}|e_1\rangle + a_{2\color{red}2}|e_2\rangle $$

(Note: In your question, you originally incorrectly iterated over j in the sum.)

BTW, treating bras as row vectors, doing $\hat e_i^T M$ picks out a row, and $\hat e_i^T M \hat e_j$ picks out the $m_{ij}$ component:

$$ \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \\ \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} = m_{12} $$

So, $a_{ij} = \langle e_i|A|e_j\rangle$.


The form of an operator expressed under a basis vectors is a matrix. The "components of the operator" are the elements of the matrix.


Matrix Elements of an Operator

What you are describing is what is called the matrix components of an operator, not its components. A linear operator on a $d$ dimensional Hilbert space can be completely and consistently described by specifying the $d\times d$ inner-products $\langle e_i\vert A\vert e_j\rangle$ in any complete orthonormal basis $\{e_i\}$ where $i,j=1,2,...,d$. These numbers can be seen as forming a $d\times d$ matrix with its $ij^\mathrm{th}$ component $A_{ij}$ given by $\langle e_i\vert A\vert e_j\rangle$. So, naturally, the $\langle e_i\vert A\vert e_j\rangle$s are called the matrix elements of the operator $A$. Once you specify these matrix elements in a given basis, of course, you can represent the operator in a basis-independent way as $A=\sum_{i,j}A_{ij} \vert i\rangle\langle j\vert$.

I don't think there is a context-independent way of ascribing to them a geometric meaning.

Components of an Operator

Interestingly, there is a sense in which you can talk about the components of an operator defined over the Hilbert space if you want to.

The set of all operators on a Hilbert space of dimension $d$ form a $2d^2$ dimensional vector space of their own over the field of complex numbers with the inner-product of two operators $A, B$ (who are vectors in this vector space of operators) is given by the trace $\mathrm{Tr}(B^\dagger A)$. Thus, you can choose a basis set of $2d^2$ orthonormal operators $O_i$ for this vector space of operators and you can completely and consistently describe an operator $A$ by specifying its expansion in the basis set as $A=\sum_i \mathrm{Tr}(O_i^\dagger A)O_i$. Now, you can treat the $\mathrm{Tr}(O_i^\dagger A)$ as the components of the operator $A$ in the basis $\{O_i\}$.


You surely were taught that $A_{ij}\equiv \langle e_i|A|e_j\rangle$, as well as $$ {\mathbb I}= \sum_{i=1}^n |e_i\rangle \langle e_i| ~~. $$ Consequently, $$ {\mathbb I} A|e_j\rangle = \sum_{i=1}^n |e_i\rangle A_{ij}, $$ so your expression which I corrected,
$$ A|e_j\rangle = \sum_{i=1}^n A_{ij}|e_i\rangle ~.$$

The components (matrix elements in a basis) are numbers which may be placed anywhere w.r.t. vectors. They matrix-multiply all components of a vector to give you all components of the linear image vector.

It is just plain linear algebra, with hardly any more meaning than the standard operation of a matrix on vectors: rotations, reflections, shearings and scalings, extended to the complex domain.


What are these "components" of operators that I am getting? Is there a geometrical interpretation for this?

You have already written what you get when the operator $A$ operates on a basis vector $|e_j\rangle$:

$$A|e_j\rangle = \sum_{i=1}^n A_{ij}|e_i\rangle$$

We can represent the operator $A$ by its matrix (with matrix elements $A_{kl}=\langle e_k|A|e_l\rangle$) and the basis states $|e_k\rangle$ by column vectors (with a $1$ in row $k$, and $0$ elsewhere). Using matrix multiplication we get:

$$\begin{pmatrix} A_{11} & A_{12} & \cdots & \color{red}{A_{1j}} & \cdots & A_{1n} \\ A_{21} & A_{22} & \cdots & \color{red}{A_{2j}} & \cdots & A_{2n} \\ \vdots & \vdots & & \color{red}{\vdots} & & \vdots \\ A_{j1} & A_{j2} & \cdots & \color{red}{A_{jj}} & \cdots & A_{jn} \\ \vdots & \vdots & & \color{red}{\vdots} & & \vdots \\ A_{n1} & A_{n2} & \cdots & \color{red}{A_{nj}} & \cdots & A_{nn} \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ \vdots \\ \color{red}{1} \\ \vdots \\ 0 \end{pmatrix}= \begin{pmatrix} \color{red}{A_{1j} \\ A_{2j} \\ \vdots \\ A_{jj} \\ \vdots \\ A_{nj}} \end{pmatrix} $$

So you have a geometrical interpretation of the $j$th column of the matrix elements of $A$. They are the vector components of $A|e_j\rangle$ in the basis $\{|e_1\rangle, ..., |e_n\rangle\}$.


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