Bra-Ket Notation

Bra-ket notation is Dirac’s compact notation for vectors, dual vectors, inner products, outer products, operators, and matrix elements. It is notation, not a separate algebra. Every valid bra-ket expression has an underlying linear-algebra meaning.

The canonical translation is:

$$\begin{array}{ccl} |\psi\rangle&:&\text{vector},\\ \langle\phi|&:&\text{dual vector},\\ \langle\phi|\psi\rangle&:&\text{scalar},\\ |\psi\rangle\langle\phi|&:&\text{operator},\\ \langle\phi|A|\psi\rangle&:&\text{scalar matrix element}. \end{array}$$

Use Dirac Notation as Linear Algebra for the mathematical construction and Representation Translation Table for the bra-ket, matrix, and wavefunction dictionary.

Kets

A ket $|\psi\rangle$ denotes a vector in a complex Hilbert space $\mathcal H$. The symbol inside the ket is a label, not automatically a number. It may name:

• a state, such as $|\psi\rangle$;
• a basis vector, such as $|n\rangle$;
• a set of quantum numbers, such as $|n\ell m\rangle$;
• a subsystem value, such as $|0\rangle_A$;
• a generalized eigenvalue, such as $|x\rangle$ or $|p\rangle$.

Linear combinations are ordinary vector sums:

$$|\chi\rangle =a|\psi\rangle+b|\phi\rangle.$$

Coefficients multiply kets from the left by convention. Writing $|\psi\rangle a$ is avoided because operators also act from the left and the ordering would become ambiguous outside scalar algebra.

Bras

The bra associated with $|\psi\rangle$ is its adjoint:

$$\langle\psi| =\left(|\psi\rangle\right)^\dagger.$$

If

$$|\psi\rangle =\sum_n c_n|n\rangle,$$

then

$$\langle\psi| =\sum_n c_n^*\langle n|.$$

Taking an adjoint reverses the order of products:

$$\left(A|\psi\rangle\right)^\dagger =\langle\psi|A^\dagger.$$

In a chosen orthonormal basis, kets may be represented by columns and bras by conjugate-transposed rows. That representation is useful, but the bra is not defined as a row of numbers independently of the inner product and basis.

Physics Inner-Product Convention

Use the physics convention:

$$\langle\phi|\psi\rangle.$$

It is conjugate-linear in $\phi$ and linear in $\psi$. Thus

$$\begin{aligned} \langle a\phi_1+b\phi_2|\psi\rangle &= a^*\langle\phi_1|\psi\rangle +b^*\langle\phi_2|\psi\rangle,\\ \langle\phi|a\psi_1+b\psi_2\rangle &= a\langle\phi|\psi_1\rangle +b\langle\phi|\psi_2\rangle. \end{aligned}$$

Conjugate symmetry gives

$$\langle\phi|\psi\rangle =\langle\psi|\phi\rangle^*.$$

The norm is

$$\|\psi\| =\sqrt{\langle\psi|\psi\rangle}.$$

The detailed comparison with the mathematics convention belongs to Inner Product Conventions.

States, Vectors, and Rays

A normalized ket may represent a pure state, but the physical state is a ray:

$$|\psi\rangle \sim e^{i\alpha}|\psi\rangle.$$

An overall nonzero rescaling also leaves the ray unchanged before normalization. Once a normalized representative is chosen, only a phase freedom remains.

Relative phases are physical. The states

$$\frac{|0\rangle+|1\rangle}{\sqrt2} \quad\text{and}\quad \frac{|0\rangle-|1\rangle}{\sqrt2}$$

are not related by one overall phase and generally give different interference probabilities.

See Rays and Global Phase for the canonical physical treatment.

Inner Products as Amplitudes

The scalar $\langle\phi|\psi\rangle$ is always an inner product when both vectors lie in the Hilbert space. It becomes a transition or measurement amplitude only after a physical question is specified.

If $|\psi\rangle$ is normalized and a projective measurement asks whether the state lies along normalized $|\phi\rangle$, then

$$p(\phi|\psi) =|\langle\phi|\psi\rangle|^2.$$

The vertical bar in $p(\phi|\psi)$ denotes conditional notation; the vertical bars in $|\langle\phi|\psi\rangle|$ denote absolute value; the vertical bars in $|\psi\rangle$ delimit a ket. Context and paired delimiters distinguish these uses.

For a degenerate outcome represented by projector $P_a$, the probability is

$$p(a|\psi) =\langle\psi|P_a|\psi\rangle,$$

not necessarily the squared overlap with one vector.

Outer Products

The outer product

$$|\psi\rangle\langle\phi|$$

is an operator. It acts on $|\chi\rangle$ as

$$\left( |\psi\rangle\langle\phi| \right)|\chi\rangle =|\psi\rangle \langle\phi|\chi\rangle.$$

The order matters. In general,

$$|\psi\rangle\langle\phi| \ne |\phi\rangle\langle\psi|.$$

Their adjoints are related by

$$\left( |\psi\rangle\langle\phi| \right)^\dagger =|\phi\rangle\langle\psi|.$$

An outer product is not an inner product:

$$\underbrace{\langle\phi|\psi\rangle}_{\text{scalar}}, \qquad \underbrace{|\psi\rangle\langle\phi|}_{\text{operator}}.$$

Projectors

If $|\psi\rangle$ is normalized, then

$$P_\psi=|\psi\rangle\langle\psi|$$

is the orthogonal projector onto the one-dimensional subspace spanned by $|\psi\rangle$. It satisfies

$$P_\psi^\dagger=P_\psi, \qquad P_\psi^2=P_\psi.$$

The projector is invariant under global phase:

$$\left(e^{i\alpha}|\psi\rangle\right) \left(e^{i\alpha}|\psi\rangle\right)^\dagger =|\psi\rangle\langle\psi|.$$

For an orthonormal set spanning a subspace $\mathcal S$,

$$P_{\mathcal S} =\sum_{a\in\mathcal S}|a\rangle\langle a|.$$

The result does not depend on which orthonormal basis is chosen inside $\mathcal S$. See Projectors for the canonical operator treatment.

Discrete Basis Expansions

For an orthonormal basis $\{|n\rangle\}$,

$$\langle m|n\rangle=\delta_{mn}, \qquad \sum_n|n\rangle\langle n|=I.$$

Inserting the identity gives

$$\begin{aligned} |\psi\rangle &=\sum_n|n\rangle\langle n|\psi\rangle\\ &=\sum_n c_n|n\rangle, \qquad c_n=\langle n|\psi\rangle. \end{aligned}$$

The bra expansion is

$$\langle\psi| =\sum_n c_n^*\langle n|.$$

For a normalized state,

$$\sum_n|c_n|^2=1.$$

Completeness must not be confused with normalization: the basis resolves the identity, while the state has unit norm.

Continuous Bases

Position and momentum “bases” are generalized spectral bases. Formally,

$$\langle x|x'\rangle=\delta(x-x'), \qquad \int dx\,|x\rangle\langle x|=I.$$

The position-space wavefunction is

$$\psi(x)=\langle x|\psi\rangle,$$

and the state is reconstructed formally as

$$|\psi\rangle =\int dx\,|x\rangle\psi(x).$$

The ket $|x\rangle$ is delta normalized and is not an ordinary normalizable Hilbert-space vector. These formulas are made precise through the spectral theorem or a rigged Hilbert space. Use Continuous Spectra and Rigged Hilbert Spaces: A First Look when that distinction matters.

Operators and Matrix Elements

Operators act on kets from the left:

$$A|\psi\rangle.$$

The product $AB$ means first apply $B$, then $A$:

$$AB|\psi\rangle=A(B|\psi\rangle).$$

In basis $\{|n\rangle\}$, the matrix elements are

$$A_{mn}=\langle m|A|n\rangle.$$

The operator can be expanded as

$$A =\sum_{mn}|m\rangle \langle m|A|n\rangle \langle n|$$

for a finite-dimensional or suitably controlled discrete setting.

The expectation value in normalized $|\psi\rangle$ is

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

The subscript on $\langle A\rangle_\psi$ labels the state; the brackets here mean expectation value rather than a bra-ket pair.

Changes of Basis

Suppose $|e_n\rangle$ and $|f_\alpha\rangle$ are orthonormal bases. Their overlap matrix is

$$U_{\alpha n} =\langle f_\alpha|e_n\rangle.$$

Components transform as

$$\langle f_\alpha|\psi\rangle =\sum_n \langle f_\alpha|e_n\rangle \langle e_n|\psi\rangle.$$

The abstract ket has not changed. Only its components in the chosen basis have changed. A physical unitary transformation of the state and a passive change of basis can use similar matrices; the surrounding statement must distinguish them.

Composite Systems

Use subsystem labels when ambiguity is possible:

$$|\psi\rangle_A\otimes|\phi\rangle_B.$$

The abbreviated form $|\psi\rangle_A|\phi\rangle_B$ is allowed when the tensor product is unambiguous, but an ordinary product of kets is not defined within one Hilbert space.

An operator local to subsystem $A$ is

$$A\otimes I_B.$$

Suppressing $I_B$ is acceptable only after the subsystem action has been declared. Tensor-factor order follows Tensor Product Ordering.

For product bases,

$$|i,j\rangle \equiv |i\rangle_A\otimes|j\rangle_B$$

only after that abbreviation and ordering have been declared.

Common Labels and Ambiguities

• $|n\rangle$ usually labels a discrete basis state, but the meaning of $n$ must be stated.
• $|x\rangle$ and $|p\rangle$ denote generalized eigenkets.
• $|0\rangle$ may mean a qubit basis state, oscillator ground state, or Fock vacuum.
• $|\Omega\rangle$ often denotes a vacuum or distinguished ground state in many-body and field-theory contexts.
• $|\psi(t)\rangle$ is a time-dependent ket in the Schrödinger picture.
• $|a,\lambda\rangle$ uses multiple labels; commas do not imply tensor products.

Never infer the physical system from the typography alone.

Common Mistakes

• Treating a bra as an unconjugated transpose.
• Forgetting that adjoints reverse operator order.
• Confusing the scalar $\langle\phi|\psi\rangle$ with the operator $|\psi\rangle\langle\phi|$.
• Reading every overlap as a probability amplitude without specifying a measurement.
• Treating $|\psi\rangle$ and $e^{i\alpha}|\psi\rangle$ as different pure states while ignoring physically meaningful relative phases.
• Inserting $\sum_n|n\rangle\langle n|=I$ for an incomplete set.
• Treating continuous kets as normalizable vectors.
• Omitting tensor-product order or subsystem identity factors when ambiguity remains.
• Reading $AB|\psi\rangle$ as though $A$ acts first.
• Using the same label $|0\rangle$ for different Hilbert spaces without a subsystem or context declaration.

Exercises

Exercise 1: Construct the bra

Let

$$|\psi\rangle =\frac{1}{\sqrt6} \left( |0\rangle+i|1\rangle-2|2\rangle \right).$$

Write $\langle\psi|$ in the same orthonormal basis and verify normalization.

Solution

Taking the adjoint conjugates coefficients:

$$\langle\psi| =\frac{1}{\sqrt6} \left( \langle0|-i\langle1|-2\langle2| \right).$$

Orthonormality gives

$$\langle\psi|\psi\rangle =\frac{1}{6} \left(1+1+4\right) =1.$$

Exercise 2: Outer-product matrix

In the ordered basis $\{|0\rangle,|1\rangle\}$, let

$$|\psi\rangle =\frac{|0\rangle+i|1\rangle}{\sqrt2}, \qquad |\phi\rangle=|0\rangle.$$

Find the matrix of $|\psi\rangle\langle\phi|$ and its adjoint.

Solution

The column and row representations are

$$|\psi\rangle =\frac{1}{\sqrt2} \begin{pmatrix} 1\\ i \end{pmatrix}, \qquad \langle\phi| = \begin{pmatrix} 1&0 \end{pmatrix}.$$

Therefore

$$|\psi\rangle\langle\phi| =\frac{1}{\sqrt2} \begin{pmatrix} 1&0\\ i&0 \end{pmatrix}.$$

Its adjoint is

$$|\phi\rangle\langle\psi| =\frac{1}{\sqrt2} \begin{pmatrix} 1&-i\\ 0&0 \end{pmatrix}.$$

Exercise 3: Insert a resolution of identity

Show that

$$\langle\phi|A|\psi\rangle =\sum_{mn} \langle\phi|m\rangle \langle m|A|n\rangle \langle n|\psi\rangle$$

for a complete orthonormal discrete basis.

Solution

Insert the identity on both sides of $A$:

$$\displaystyle I=\sum_n|n\rangle\langle n|.$$

Then

$$\begin{aligned} \langle\phi|A|\psi\rangle &=\langle\phi|IAI|\psi\rangle\\ &= \sum_{mn} \langle\phi|m\rangle \langle m|A|n\rangle \langle n|\psi\rangle. \end{aligned}$$

This is ordinary matrix multiplication written without choosing coordinate columns at the outset.

Exercise 4: Global phase and a projector

Let $|\psi'\rangle=e^{i\alpha}|\psi\rangle$. Show that the rank-one projector and all expectation values are unchanged.

Solution

The bra transforms as

$$\langle\psi'| =e^{-i\alpha}\langle\psi|.$$

Hence

$$|\psi'\rangle\langle\psi'| =e^{i\alpha}e^{-i\alpha} |\psi\rangle\langle\psi| =|\psi\rangle\langle\psi|.$$

For any operator $A$ whose expectation is defined,

$$\langle\psi'|A|\psi'\rangle =e^{-i\alpha}e^{i\alpha} \langle\psi|A|\psi\rangle =\langle\psi|A|\psi\rangle.$$

References

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