KroneckerDelta[m, n], displayed as $\mathrm{\δ\_\_m,n}$, is a computational representation for the Kronecker Delta, that is: it returns 1 or 0, according to whether the indices are equal or different, or so assumed using assuming. When equality cannot be proved or disproved KroneckerDelta[m, n] returns unevaluated, echoing the input.

Note: since Maple 2019, KroneckerDelta[m, n] is not implemented as a tensor (of type Physics:-Library:-PhysicsType:-Tensor), even if $m$ and $n$ are letters representing spacetime or space tensor indices, unless one of such indices is contravariant, preceded with ~, as in ~n, while the other remains covariant (in which case KroneckerDelta[m, ~n] is automatically transformed into the spacetime metric g_[m, n] which is a tensor). Consequently, $\mathrm{\δ\_\_m,m}$ with the two indices covariant, returns the number 1 in all cases, and Einstein's sum rule for repeated indices is not applied to $\mathrm{\δ\_\_m,n}$ even when $m$ or $n$ appear repeated in a product of an algebraic expression.

If, on the other hand, you prefer to work with KroneckerDelta[m, n] as a tensor, for example in quantum mechanics by setting the spacetime to be Euclidean and of dimension 3 (something you can accomplish using Setup), you can set a macro, or Define a tensor representing the Kronecker $\mathrm{\delta }$ tensor. See the example at the end of the Examples section.

The %KroneckerDelta command is the inert form of KroneckerDelta, so it represents the same mathematical operation but without performing it. To perform the operation, use value.

Some automatic checking and normalization are carried out each time you enter KroneckerDelta[...]. The checking is concerned with possible syntax errors. The automatic normalization takes into account the symmetry of KroneckerDelta[m,n] with respect to interchanging the positions of the indices m and n.

The %KroneckerDelta command is the inert form of KroneckerDelta, so it represents the same thing but entering it does not result in performing any computation. To perform the related computations as if %KroneckerDelta were KroneckerDelta, use value.

$\mathrm{with}\left(\mathrm{Physics}\right)\:$

$\mathrm{Setup}\left(\mathrm{mathematicalnotation}\=\mathrm{true}\right)$

$\left[\mathrm{mathematicalnotation}=\mathrm{true}\right]$

${\mathrm{KroneckerDelta}}_{2\,2}\={\mathrm{KroneckerDelta}}_{m\,m}$

$1=1$

${\mathrm{KroneckerDelta}}_{2\,3}$

$0$

${\mathrm{KroneckerDelta}}_{m\,n}$

${\mathrm{\delta }}_{m,n}$

$assumingm\=n$

$1$

$assumingm\ne n$

$0$

${\mathrm{KroneckerDelta}}_{m\,n}-{\mathrm{KroneckerDelta}}_{n\,m}$

$0$

$\mathrm{Setup}\left(\mathrm{spacetimeindices}\,\mathrm{dimension}\right)$

$\left[\mathrm{dimension}=4\,\mathrm{spacetimeindices}=\mathrm{greek}\right]$

${\mathrm{KroneckerDelta}}_{\mathrm{\μ}\,\mathrm{\μ}}$

$1$

${\mathrm{\%KroneckerDelta}}_{\mathrm{\μ}\,\mathrm{\μ}}$

${\mathrm{\%KroneckerDelta}}_{\mathrm{\mu },\mathrm{\mu }}$

$\mathrm{value}\left(\right)$

$1$

${\mathrm{KroneckerDelta}}_{\mathrm{\μ}\,\mathrm{\ν}}$

${\mathrm{\delta }}_{\mathrm{\mu },\mathrm{\nu }}$

$\mathrm{type}\left(\,\mathrm{Physics}:-\mathrm{Library}:-\mathrm{PhysicsType}:-\mathrm{Tensor}\right)$

$\mathrm{false}$

${\mathrm{KroneckerDelta}}_{\mathrm{\μ}\,\mathrm{`~nu`}}\={\mathrm{g\_}}_{\mathrm{\μ}\,\mathrm{`~nu`}}$

${\mathrm{g\_}}_{\mathrm{\mu }\phantom{\mathrm{\nu }}}^{\phantom{\mathrm{\mu }}\mathrm{\nu }}={\mathrm{g\_}}_{\mathrm{\mu }\phantom{\mathrm{\nu }}}^{\phantom{\mathrm{\mu }}\mathrm{\nu }}$

$\mathrm{lprint}\left(\right)$

Physics:-g_[mu,~nu] = Physics:-g_[mu,~nu]

$\genfrac{}{}{0ex}{}{\phantom{\mathrm{\μ}\=\mathrm{\ν}}}{}|\genfrac{}{}{0ex}{}{\phantom{}}{\mathrm{\μ}\=\mathrm{\ν}}$

$4=4$

$assuming\mathrm{\μ}\=\mathrm{\ν}$

$4=4$

$\mathrm{Setup}\left(\mathrm{metric}\=\mathrm{Euclidean}\,\mathrm{dimension}\=3\,\mathrm{spacetimeindices}\=\mathrm{lowercaselatin}\right)$

$\mathrm{The\; dimension\; and\; signature\; of\; the\; tensor\; space\; are\; set\; to}\left[3\,\left(\mathrm{-\; -\; +}\right)\right]$

$\mathrm{The\; Euclidean\; metric\; in\; cartesian\; coordinates}$

$\mathrm{Changing\; the\; signature\; of\; the\; tensor\; spacetime\; to:}\mathrm{+\; +\; +}$

$\left[\mathrm{dimension}=3\,\mathrm{metric}=\left\{\left(1\,1\right)=1\,\left(2\,2\right)=1\,\left(3\,3\right)=1\right\}\,\mathrm{spacetimeindices}=\mathrm{lowercaselatin}\right]$

$\mathrm{Setup}\left(\mathrm{quantumoperators}\=\left\{p\,q\right\}\,\mathrm{tensors}\=\left\{q_j\,p_k\right\}\,\mathrm{metric}\=\mathrm{Euclidean}\,\mathrm{dimension}\=3\,\mathrm{spacetimeindices}\=\mathrm{lowercase}\right)$

$\mathrm{The\; Euclidean\; metric\; in\; cartesian\; coordinates}$

$\left[\mathrm{dimension}=3\,\mathrm{metric}=\left\{\left(1\,1\right)=1\,\left(2\,2\right)=1\,\left(3\,3\right)=1\right\}\,\mathrm{quantumoperators}=\left\{p\,q\right\}\,\mathrm{spacetimeindices}=\mathrm{lowercaselatin}\,\mathrm{tensors}=\left\{{\mathrm{\gamma }}_a\,{\mathrm{\sigma }}_a\,{\partial }_a\,g_{a,b}\,p_k\,q_j\,{\mathrm{\epsilon }}_{a,b,c}\right\}\right]$

$\mathrm{Setup}\left(\mathrm{\%Commutator}\left(p_j\,q_k\right)\={\mathrm{KroneckerDelta}}_{j\,k}\right)$

$\mathit{Warning,\; since\; Maple\; 2019\; KroneckerDelta\; is\; not\; implemented\; as\; a\; tensor.\; Thus,\; the\; right-hand\; side\; of\; the\; algebra\; rule}\left(\mathrm{`?`}\mathbf{=}{\mathbf{\delta }}_{\mathit{j}\mathbf{,}\mathit{k}}\right)\mathit{has\; for\; free\; tensor\; indices}\mathit{\varnothing }\mathit{,\; while\; the\; left-hand\; side\; has}\left\{\mathit{j}\,\mathit{k}\right\}\mathit{,\; of\; which}\left\{\mathit{j}\,\mathit{k}\right\}\mathit{appear\; in}\left\{{\mathbf{\delta }}_{\mathit{j}\mathbf{,}\mathit{k}}\right\}\mathit{.\; To\; resolve\; this\; problem\; you\; can:\; 1.\; Use\; the\; spacetime\; metric\; g\_,\; or\; the\; space\; metric\; gamma\_,\; instead\; of\; KroneckerDelta.\; If\; so,\; check\; the\; value\; of:}\mathit{Setup}\left(\mathit{signature}\right)\mathit{.\; 2.\; Enter:}\mathit{macro}\left(\mathit{KroneckerDelta}\mathbf{=}\mathit{g\_}\right)\mathit{;\; or\; use\; gamma\_\; instead\; of\; g\_.\; 3.\; Define\; a\; Kronecker}\mathbf{\delta }\mathit{tensor.\; In\; the\; following,\; type\; \text{'}delta\text{'}\; on\; the\; left\; and\; \text{'}KroneckerDelta\text{'}\; on\; the\; right,\; or\; copy\; and\; paste:}\mathit{Define}\left({\mathbf{\delta }}_{\mathit{a}\mathbf{,}\mathit{b}}\mathbf{=}\mathit{Matrix}\left(\mathbf{3}\,\mathbf{3}\,\left(\mathit{j}\,\mathit{k}\right)\mathbf{\mapsto }{\mathbf{\delta }}_{\mathit{j}\mathbf{,}\mathit{k}}\right)\right)\mathit{.\; After\; any\; of\; the\; above\; it\; is\; recommended\; that\; you\; set\; this\; algebra\; rule\; again,\; with\; the\; corrected\; right-hand\; side,\; to\; avoid\; problems\; due\; to\; having\; different\; free\; indices\; on\; each\; side\; of\; the\; rule.}$

$\left[\mathrm{algebrarules}\=\left\{\mathrm{\%Commutator}\left(p_j\,q_k\right)\={\mathrm{KroneckerDelta}}_{j\,k}\right\}\right]$

$\mathrm{macro}\left(\mathrm{KroneckerDelta}\=\mathrm{g\_}\right)\:$

${\mathrm{KroneckerDelta}}_{m\,m}$

$3$

$\mathrm{macro}\left(\mathrm{KroneckerDelta}\=\mathrm{KroneckerDelta}\right)\:$

$\mathrm{Define}\left({\mathrm{\δ}}_{a\,b}\=\mathrm{Matrix}\left(3\,3\,\left(j\,k\right)\→{\mathrm{KroneckerDelta}}_{j\,k}\right)\right)$

$\mathrm{Defined\; objects\; with\; tensor\; properties}$

$\left\{{\mathrm{\gamma }}_a\,{\mathrm{\sigma }}_a\,{\partial }_a\,{\mathrm{\δ}}_{a,b}\,g_{a,b}\,p_k\,q_j\,{\mathrm{\epsilon }}_{a,b,c}\right\}$

${\mathrm{\δ}}_{\[\]}$

${\mathrm{\δ}}_{a\,b}\=\left(\left[\begin{array}{rrr}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\right)$

${\mathrm{\δ}}_{1\,1}$

$1$

$\mathrm{SumOverRepeatedIndices}\left({\mathrm{\δ}}_{m\,m}\right)$

$3$

${\mathrm{KroneckerDelta}}_{m\,m}$

$1$

$\mathrm{Setup}\left(\mathrm{\%Commutator}\left(p_j\,q_k\right)\={\mathrm{\δ}}_{j\,k}\right)$

$\left[\mathrm{algebrarules}\=\left\{\mathrm{\%Commutator}\left(p_j\,q_k\right)\={\mathrm{\δ}}_{j\,k}\right\}\right]$

$\mathrm{Matrix}\left(3\,\left(m\,n\right)\→\left(\mathrm{\%Commutator}\=\mathrm{Commutator}\right)\left(p_m\,q_n\right)\right)$

$\left[\begin{array}{ccc}\mathrm{\%Commutator}\left(p_1\,q_1\right)\=1& \mathrm{\%Commutator}\left(p_1\,q_2\right)\=0& \mathrm{\%Commutator}\left(p_1\,q_3\right)\=0\\ \mathrm{\%Commutator}\left(p_2\,q_1\right)\=0& \mathrm{\%Commutator}\left(p_2\,q_2\right)\=1& \mathrm{\%Commutator}\left(p_2\,q_3\right)\=0\\ \mathrm{\%Commutator}\left(p_3\,q_1\right)\=0& \mathrm{\%Commutator}\left(p_3\,q_2\right)\=0& \mathrm{\%Commutator}\left(p_3\,q_3\right)\=1\end{array}\right]$

$\left(\mathrm{\%Commutator}\=\mathrm{Commutator}\right)\left(p_a\,q_b\right)$

$\mathrm{\%Commutator}\left(p_a\,q_b\right)\={\mathrm{\δ}}_{a\,b}$

$\mathrm{expand}\left(\genfrac{}{}{0ex}{}{\phantom{a\=b}}{}|\genfrac{}{}{0ex}{}{\phantom{}}{a\=b}\right)$

$p_b{q}_b-q_b{p}_b={\mathrm{\δ}}_{b,b}$

$\mathrm{SumOverRepeatedIndices}\left(\right)$

$p_1{q}_1+p_2{q}_2+p_3{q}_3-q_1{p}_1-q_2{p}_2-q_3{p}_3=3$

$\mathrm{Simplify}\left(\right)$

$3=3$

The Physics[KroneckerDelta] command was updated in Maple 2019.