Dirac Notation as Linear Algebra

Dirac notation is compact linear algebra for complex inner-product spaces. A ket is a vector, a bra is a dual vector obtained from the inner product convention, an inner product is a scalar, an outer product is a linear map, and a matrix element is a coordinate of an operator.

We first work in a finite-dimensional complex inner-product space; Bra-Ket Notation gives the shared notation conventions. The final section explains what changes for continuous position labels.

Helpful background. complex vector spaces for complex coefficients and scalar multiplication; inner products for norms, conjugation, and orthogonality.

Kets Are Vectors

A ket such as

$$\lvert\psi\rangle$$

denotes a vector in a complex vector space, usually a Hilbert space $\mathcal H$. In a chosen basis $\{\lvert e_j\rangle\}$, it can be represented by coordinates:

$$\lvert\psi\rangle = \sum_j c_j\lvert e_j\rangle.$$

The ket is the abstract vector. The coordinate column

$$\begin{pmatrix} c_1\\ c_2\\ \vdots \end{pmatrix}$$

is a representation of that vector after the basis has been chosen. Changing the basis changes the coordinates, not the vector.

Bras Are Dual Vectors

A bra such as

$$\langle\phi\rvert$$

is a linear functional on kets:

$$\langle\phi\rvert:\mathcal H\to\mathbb C.$$

With the physics convention, the inner product

$$\langle\phi\vert\psi\rangle$$

is conjugate-linear in $\phi$ and linear in $\psi$. Therefore the map from kets to corresponding bras is conjugate-linear:

$$a\lvert\phi\rangle+b\lvert\chi\rangle \mapsto a^*\langle\phi\rvert+b^*\langle\chi\rvert.$$

In a finite orthonormal basis, if

$$\lvert\psi\rangle \leftrightarrow \begin{pmatrix} c_1\\ \vdots\\ c_n \end{pmatrix},$$

then

$$\langle\psi\rvert \leftrightarrow \begin{pmatrix} c_1^* & \cdots & c_n^* \end{pmatrix}.$$

This is the linear-algebra content behind the rule “take the conjugate transpose.”

Inner Products Are Scalars

The product

$$\langle\phi\vert\psi\rangle$$

is a complex number. In an orthonormal basis,

$$\lvert\psi\rangle = \sum_j c_j\lvert e_j\rangle, \qquad \lvert\phi\rangle = \sum_j d_j\lvert e_j\rangle,$$

so

$$\langle\phi\vert\psi\rangle = \sum_j d_j^*c_j.$$

When both vectors are normalized, $\langle\phi\vert\psi\rangle$ is a transition amplitude and

$$\lvert\langle\phi\vert\psi\rangle\rvert^2$$

is the corresponding squared amplitude. The Born rule gives the probability interpretation; the linear-algebra fact is that the inner product pairs a dual vector with a vector.

Outer Products Are Linear Maps

The expression

$$\lvert u\rangle\langle v\rvert$$

is not an inner product. It is a linear map from $\mathcal H$ to itself. Acting on a ket $\lvert\psi\rangle$, it gives

$$\left( \lvert u\rangle\langle v\rvert \right) \lvert\psi\rangle = \lvert u\rangle \langle v\vert\psi\rangle.$$

Thus the bra first extracts a scalar from $\lvert\psi\rangle$, and the ket $\lvert u\rangle$ supplies the output direction.

If $\lvert u\rangle$ is normalized, then

$$P_u = \lvert u\rangle\langle u\rvert$$

is a rank-one projector:

$$P_u^2=P_u, \qquad P_u^\dagger=P_u.$$

See Projectors for projection onto larger subspaces.

Operators and Matrix Elements

An operator $A$ is a linear map on the state space:

$$A:\mathcal H\to\mathcal H.$$

In an orthonormal basis $\{\lvert e_j\rangle\}$, the matrix entry of $A$ is

$$A_{ij} = \langle e_i\vert A\vert e_j\rangle.$$

This formula says: feed the basis vector $\lvert e_j\rangle$ into $A$, then extract the component along $\lvert e_i\rangle$ by applying $\langle e_i\rvert$. With the convention

$$A\lvert e_j\rangle = \sum_i A_{ij}\lvert e_i\rangle,$$

the same numbers $A_{ij}$ form the matrix of the operator in that basis.

An expectation value is a special matrix element:

$$\langle A\rangle_\psi = \langle\psi\vert A\vert\psi\rangle.$$

For normalized $\lvert\psi\rangle$, this is the expected value of the observable represented by $A$ when $A$ is self-adjoint and the relevant domain issues are under control.

Completeness and Basis Expansions

For an orthonormal basis, the identity operator can be written

$$I = \sum_j \lvert e_j\rangle\langle e_j\rvert.$$

Applying this identity to a vector gives the familiar expansion:

$$\lvert\psi\rangle = I\lvert\psi\rangle = \sum_j \lvert e_j\rangle \langle e_j\vert\psi\rangle.$$

The coefficient $\langle e_j\vert\psi\rangle$ is the $j$th coordinate of the vector in this orthonormal basis.

Continuous bases use analogous expressions such as

$$I = \int \lvert x\rangle\langle x\rvert\,dx,$$

but $\lvert x\rangle$ is a generalized eigenket, not an ordinary normalizable Hilbert-space vector. The rigorous home of this issue is distribution theory and spectral theory; the practical translation is summarized in Representation Translation Table.

Worked Example

Let

$$\lvert\psi\rangle = \frac{1}{\sqrt3} \begin{pmatrix} 1\\ 1+i \end{pmatrix}, \qquad \lvert\phi\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix}.$$

Then

$$\langle\psi\rvert = \frac{1}{\sqrt3} \begin{pmatrix} 1 & 1-i \end{pmatrix},$$

and

$$\langle\phi\vert\psi\rangle = \frac{1+i}{\sqrt3}.$$

The outer product $\lvert\psi\rangle\langle\phi\rvert$ is the matrix

$$\lvert\psi\rangle\langle\phi\rvert = \frac{1}{\sqrt3} \begin{pmatrix} 1\\ 1+i \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix} = \frac{1}{\sqrt3} \begin{pmatrix} 0 & 1\\ 0 & 1+i \end{pmatrix}.$$

It maps a vector $v=(v_1,v_2)^T$ to

$$\frac{v_2}{\sqrt3} \begin{pmatrix} 1\\ 1+i \end{pmatrix}.$$

This example displays the main distinction: $\langle\phi\vert\psi\rangle$ is a scalar, while $\lvert\psi\rangle\langle\phi\rvert$ is an operator.

Common Mistakes

• Treating $\lvert\psi\rangle$ as a column independent of a basis.
• Turning a ket into a bra without complex conjugating coefficients.
• Confusing $\langle\phi\vert\psi\rangle$ with $\lvert\psi\rangle\langle\phi\rvert$.
• Reading $\langle e_i\vert A\vert e_j\rangle$ in the wrong order.
• Forgetting that continuous kets such as $\lvert x\rangle$ are generalized objects.
• Inserting completeness relations without checking which basis or measure is being used.

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.

Exercises

1. Let

$$\lvert\psi\rangle = \frac{1}{\sqrt2} \left( \lvert0\rangle+i\lvert1\rangle \right).$$

Find $\langle\psi\rvert$.

Solution

The ket-to-bra map conjugates the coefficients:

$$\langle\psi\rvert = \frac{1}{\sqrt2} \left( \langle0\rvert-i\langle1\rvert \right).$$

2. Show that $\lvert u\rangle\langle v\rvert$ is linear in the input ket.

Solution

For input $a\lvert\psi\rangle+b\lvert\chi\rangle$,

$$\lvert u\rangle\langle v\rvert \left( a\lvert\psi\rangle+b\lvert\chi\rangle \right) = \lvert u\rangle \left( a\langle v\vert\psi\rangle +b\langle v\vert\chi\rangle \right).$$

Distributing the scalar over the output vector gives

$$a\lvert u\rangle\langle v\vert\psi\rangle + b\lvert u\rangle\langle v\vert\chi\rangle.$$

Thus the outer product is a linear map on the input ket.

3. In an orthonormal basis, let $A\lvert e_2\rangle=3\lvert e_1\rangle-i\lvert e_2\rangle$. What are $A_{12}$ and $A_{22}$?

Solution

By definition,

$$A_{i2} = \langle e_i\vert A\vert e_2\rangle.$$

The coefficient of $\lvert e_1\rangle$ is $3$, and the coefficient of $\lvert e_2\rangle$ is $-i$. Hence

$$A_{12}=3, \qquad A_{22}=-i.$$