Representation Translation Table

The same quantum object can be represented in different ways. The translation is not merely typography: states, operators, inner products, measures, and update rules must all be translated together.

For the conceptual statement that these presentations can describe the same predictions, see Equivalent Formulations. This page is the quick reference for moving between the notations used most often in Core Formalism.

Helpful background. Bases and representations explains the coordinate map; Dirac notation as linear algebra explains bras and kets.

Representations and the probability pairing

A representation is a coordinate system for the same abstract object. A correct translation preserves probabilities and expectation values.

For example, the abstract state $\lvert\psi\rangle$ may be represented by:

• a column of components $c_n=\langle n\vert\psi\rangle$ in a discrete basis;
• a position-space wavefunction $\psi(x)=\langle x\vert\psi\rangle$;
• a momentum-space wavefunction $\tilde\psi(p)=\langle p\vert\psi\rangle$;
• the rank-one density operator $\rho_\psi=\lvert\psi\rangle\langle\psi\rvert$, which removes the irrelevant global phase.

These descriptions give the same pure-state predictions. The density-operator map is a bijection from rays to rank-one projectors, rather than a linear change of vector coordinates. General mixed density operators have no vector representative on the same Hilbert space.

States and Inner Products

The continuous columns describe a scalar particle on the full real line, with measure $dx$ or $dp$. The discrete basis must be complete: a finite basis for a finite system, or a complete countable basis for $L^2(\mathbb R)$. A finite truncation of a spatial problem is an approximation, so its truncated resolution is a projector rather than the full identity.

Abstract notation	Orthonormal basis	Position representation	Momentum representation
$\lvert\psi\rangle$	column vector $c_n=\langle n\vert\psi\rangle$	wavefunction $\psi(x)=\langle x\vert\psi\rangle$	wavefunction $\tilde\psi(p)=\langle p\vert\psi\rangle$
$\langle\phi\vert\psi\rangle$	$\sum_n \phi_n^*\psi_n$	$\int \phi(x)^*\psi(x)\,dx$	$\int \tilde\phi(p)^*\tilde\psi(p)\,dp$
normalization	$\sum_n \lvert c_n\rvert^2=1$	$\int \lvert\psi(x)\rvert^2\,dx=1$	$\int \lvert\tilde\psi(p)\rvert^2\,dp=1$
completeness	$\sum_n \lvert n\rangle\langle n\rvert=I$	$\int \lvert x\rangle\langle x\rvert\,dx=I$	$\int \lvert p\rangle\langle p\rvert\,dp=I$
component extraction	$c_n=\langle n\vert\psi\rangle$	$\psi(x)=\langle x\vert\psi\rangle$	$\tilde\psi(p)=\langle p\vert\psi\rangle$

The position and momentum formulas assume the default Fourier convention. See Fourier Transform Conventions and Momentum-Space Representation for the exact normalization.

Operators and Matrix Elements

Abstract notation	Orthonormal basis	Position representation	Meaning
$A$	matrix $A_{mn}=\langle m\vert A\vert n\rangle$	differential, multiplication, or integral operator	same abstract operator in different coordinates
$A\lvert\psi\rangle$	component column $(Ac)_m=\sum_n A_{mn}c_n$	function $(A\psi)(x)$	operator acting on state
$\langle\phi\vert A\vert\psi\rangle$	$\sum_{m,n}\phi_m^*A_{mn}c_n$	$\int \phi(x)^*(A\psi)(x)\,dx$	transition matrix element
$P_\psi$	matrix of $\lvert\psi\rangle\langle\psi\rvert$	kernel $\psi(x)\psi(x')^*$	rank-one projector when $\lvert\psi\rangle$ is normalized
$H$	Hamiltonian matrix	often $-\hbar^2\nabla^2/(2m)+V(\mathbf r)$	energy observable and generator of time evolution

The position-representation Hamiltonian shown here is the common single-particle Cartesian form. It is not the universal form of every Hamiltonian. Boundary conditions, domains, coordinates, spin, gauge fields, and many-body variables can change the operator representation.

For the general rule, see Operator Representations.

Probabilities and Expectation Values

Quantity	Abstract pure-state form	Basis form	Position representation	Density-operator form
outcome probability	$\langle\psi\vert P_a\vert\psi\rangle$	$c^\dagger P_a c$	$\int \psi(x)^*(P_a\psi)(x)\,dx$	$\operatorname{Tr}(\rho P_a)$
expectation value	$\langle\psi\vert A\vert\psi\rangle$	$c^\dagger A c$	$\int \psi(x)^*(A\psi)(x)\,dx$	$\operatorname{Tr}(\rho A)$
continuous position probability	$\langle\psi\vert P_R\vert\psi\rangle$	basis-dependent matrix expression	$\int_R \lvert\psi(x)\rvert^2\,dx$	$\operatorname{Tr}(\rho P_R)$
variance	$\langle A^2\rangle-\langle A\rangle^2$	$c^\dagger A^2c-(c^\dagger Ac)^2$	integral version of the same expression	$\operatorname{Tr}(\rho A^2)-\operatorname{Tr}(\rho A)^2$

For unbounded $A$, the operator-action expectation requires $\psi\in D(A)$. A finite first spectral moment can exist under weaker conditions; a finite second moment is $\|A\psi\|^2$ and need not require $\psi\in D(A^2)$. Interpret the displayed second-moment shorthand spectrally when necessary. A mixed-state operator-product trace also needs its convergence and domain conditions.

The symbol $P_R$ denotes the projector onto a spatial region $R$ in the ideal position measurement. The continuous-position probability formula is a special case of the Born rule for continuous spectra.

See Born Rule, Born Rule for Continuous Spectra, Expectation Values, and Trace Rule for Expectation Values.

Density Operators and Kernels

Abstract density-operator notation	Orthonormal basis	Position representation	Comment
$\rho$	matrix $\rho_{mn}=\langle m\vert\rho\vert n\rangle$	kernel $\rho(x,x')=\langle x\vert\rho\vert x'\rangle$	same state in different representations
$\rho_\psi=\lvert\psi\rangle\langle\psi\rvert$	$\rho_{mn}=c_m c_n^*$	$\rho(x,x')=\psi(x)\psi(x')^*$	pure-state density operator
$\operatorname{Tr}\rho=1$	$\sum_n \rho_{nn}=1$	$\int \rho(x,x)\,dx=1$	trace-one condition
$\rho_A=\operatorname{Tr}_B\rho_{AB}$	sum over the $B$ index	integral or sum over discarded variables	reduced state for subsystem $A$

For a positive trace-class state, use its spectral expansion to define the position density: if $\rho=\sum_jp_j|\psi_j\rangle\langle\psi_j|$, then $\rho(x,x)=\sum_jp_j|\psi_j(x)|^2$ almost everywhere. An arbitrary $L^2$ kernel representative need not have a meaningful diagonal; the spectral version gives the physical density. The full kernel $\rho(x,x')$ contains off-diagonal coherence information in that representation.

Dynamics

Abstract equation	Matrix representation	Position representation	Density-operator representation
$i\hbar\,d\lvert\psi\rangle/dt=H\lvert\psi\rangle$	$i\hbar\,\dot c=Hc$	$i\hbar\,\partial_t\psi(x,t)=H_x\psi(x,t)$	$i\hbar\,\dot\rho=[H,\rho]$
$\lvert\psi(t)\rangle=U(t)\lvert\psi(0)\rangle$	$c(t)=U(t)c(0)$	$\psi(x,t)=\int K(x,t;x',0)\psi(x',0)\,dx'$	$\rho(t)=U(t)\rho(0)U(t)^\dagger$
$U(t)=e^{-iHt/\hbar}$	matrix exponential or spectral decomposition	propagator kernel or differential equation solution	conjugation by $U(t)$

The exponential row assumes a time-independent self-adjoint $H$. Strong differential equations require the appropriate state domains and differentiability; unitary evolution itself is defined for every Hilbert-space state. Integral kernels can be distributions and must then be interpreted through their action on states.

Here $K(x,t;x',0)=\langle x\vert U(t)\vert x'\rangle$ is the propagator kernel. Quantum Dynamics and Formulations develops kernel methods.

Change of Basis

Suppose the old orthonormal basis is $\{\lvert e_n\rangle\}$ and the new orthonormal basis is $\{\lvert f_a\rangle\}$. With the convention

$$S_{an} = \langle f_a\vert e_n\rangle,$$

state components transform as

$$d=Sc,$$

and operator matrices transform as

$$A_f = S A_e S^\dagger.$$

The formulas must be used together. If the state column is transformed but the operator matrix is not, probabilities and expectation values generally change. See Change of Basis for the derivation and a two-level example.

Worked translation with a complex qubit

In the $Z$ basis, prepare the $+1$ eigenstate of $Y$:

$$c=\frac1{\sqrt2}\begin{pmatrix}1\\i\end{pmatrix}, \qquad Y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}.$$

Change to the $X$ basis using

$$S=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$

The new column and observable matrix are

$$d=Sc=\frac12\begin{pmatrix}1+i\\1-i\end{pmatrix}, \qquad Y_f=SYS^\dagger=-Y.$$

Since $Yc=c$ and $Y_fd=d$, both calculations give expectation $+1$. Equivalently, $\rho_f=dd^\dagger$ gives $\operatorname{Tr}(\rho_fY_f)=1$. The $+1$ outcome projector also changes to $(I+Y_f)/2=(I-Y)/2$, so its probability remains one.

The minus sign in $Y_f$ describes a coordinate change, not a reversal of the physical spin.

Check the translated observable

What would happen if you transformed $c$ to $d$ but used the original matrix $Y$?

Solution

Because $Yd=-d$, the inconsistent calculation gives $d^\dagger Yd=-1$. It represents a different physical observable in the new coordinates. Transforming both the state and the observable restores the original prediction $+1$.

For other useful checks, see Common checks and sanity tests.

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• 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.