Expressions involving sums with KroneckerDelta indices contracted or with Projectors in the summand, and integrals involving Dirac functions, are simplifiable by using Simplify.

Consider a basis, labeled $A$, whose dimension is $M+1$.

$\mathrm{Setup}\left(\mathrm{basisdim}=\left\{A=0..M\right\}\right)$

$\mathrm{*\; Partial\; match\; of\; \text{'}}\mathrm{basisdim}\mathrm{\text{'}\; against\; keyword\; \text{'}}\mathrm{quantumbasisdimension}\text{'}$

$\mathrm{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}$

$\left[\mathrm{quantumbasisdimension}=\left\{A=0..M\right\}\right]$

The projector onto this basis is:

$\mathrm{Projector}\left(\mathrm{Ket}\left(A\,j\right)\right)$

$\sum _{j=0}^M\left|A_j\right.⟩\left.⟨A_j\right|$

So this expression,

$\mathrm{Sum}\left(\mathrm{Bracket}\left(\mathrm{Bra}\left(\mathrm{\Psi }\,n\right)\,\mathrm{Ket}\left(A\,j\right)\right)\mathrm{Bracket}\left(\mathrm{Bra}\left(A\,j\right)\,\mathrm{Ket}\left(\mathrm{\Psi }\,m\right)\right)\,j=0..M\right)$

$\sum _{j=0}^M⟨{\mathrm{\Psi }}_n|A_j⟩⟨A_j|{\mathrm{\Psi }}_m⟩$

where you see the projector inserted in the Bracket, can be simplified to:

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

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

Expressions involving Sums with KroneckerDelta in the summand, provided that the indices in KroneckerDelta are inside the summation range (you can use assuming to enforce that), are also simplifiable.

$\mathrm{Sum}\left(\mathrm{Sum}\left(\mathrm{Sum}\left(\mathrm{Sum}\left(\mathrm{conjugate}\left(a\left[n,i\right]\right)a\left[i,m\right]\mathrm{KroneckerDelta}\left[i,j\right]\mathrm{KroneckerDelta}\left[m,n\right]\,j=0..M\right)\,i=0..M\right)\,n=0..M\right)\,m=0..M\right)$

$\sum _{m=0}^M\sum _{n=0}^M\sum _{i=0}^M\sum _{j=0}^M\stackrel{\&conjugate0;}{a_{n,i}}{a}_{i,m}{\mathrm{\delta }}_{i,j}{\mathrm{\delta }}_{m,n}$

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

$\sum _{m=0}^M\sum _{i=0}^M\stackrel{\&conjugate0;}{a_{m,i}}{a}_{i,m}$

In the following example, both simplifications are used together.

$\mathrm{Sum}\left(\mathrm{Sum}\left(\mathrm{Sum}\left(\mathrm{Sum}\left(\mathrm{conjugate}\left(a\left[n,i\right]\right)a\left[i,m\right]\mathrm{Bracket}\left(\mathrm{Bra}\left(\mathrm{\Psi }\,n\right)\,\mathrm{Ket}\left(A\,j\right)\right)\mathrm{Bracket}\left(\mathrm{Bra}\left(A\,j\right)\,\mathrm{Ket}\left(\mathrm{\Psi }\,m\right)\right)\,j=0..M\right)\,i=0..M\right)\,n=0..M\right)\,m=0..M\right)$

$\sum _{m=0}^M\sum _{n=0}^M\sum _{i=0}^M\sum _{j=0}^M\stackrel{\&conjugate0;}{a_{n,i}}{a}_{i,m}⟨{\mathrm{\Psi }}_n|A_j⟩⟨A_j|{\mathrm{\Psi }}_m⟩$

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

$\sum _{m=0}^M\sum _{i=0}^M\stackrel{\&conjugate0;}{a_{m,i}}{a}_{i,m}$

Two different kinds of simplifications are available: a normalization of products involving noncommutative operands some of which commute between themselves (that have Commutator equal to zero), and a simplification taking into account (Anti)Commutator rules (that have (Anti)Commutator not equal to zero). These simplifications are frequently required, for example, when deriving the Commutator algebra of problems involving angular momentum or Annihilation/Creation operators.

1. Consider the angular momentum operators $\|L\|$, $\mathrm{Lx},\mathrm{Ly}$, and $\mathrm{Lz}$ in quantum mechanics. Verify that the Commutator of $\|L\|\^2$ with any of the components of $L$ is equal to zero (see for instance Chapter VI of the reference, below). For that purpose, a 3-D vector-quantum-operator representation of $L$ is constructed with the Vectors package (vectorpostfix identifier is '_'), setting $L$, $\mathrm{Lx},\mathrm{Ly},\mathrm{Lz}$, $r$, and $p$, as well as their components, as quantum operators.

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

$\left[\mathrm{\&x}\,\mathrm{`+`}\,\mathrm{`.`}\,\mathrm{Assume}\,\mathrm{ChangeBasis}\,\mathrm{ChangeCoordinates}\,\mathrm{CompactDisplay}\,\mathrm{Component}\,\mathrm{Curl}\,\mathrm{DirectionalDiff}\,\mathrm{Divergence}\,\mathrm{Gradient}\,\mathrm{Identify}\,\mathrm{Laplacian}\,\nabla \,\mathrm{Norm}\,\mathrm{ParametrizeCurve}\,\mathrm{ParametrizeSurface}\,\mathrm{ParametrizeVolume}\,\mathrm{Setup}\,\mathrm{Simplify}\,\mathrm{`^`}\,\mathrm{diff}\,\mathrm{int}\right]$

$\mathrm{Setup}\left(\mathrm{quantumoperators}=\left\{\mathrm{L\_}\,\mathrm{Lx}\,\mathrm{Ly}\,\mathrm{Lz}\,\mathrm{p\_}\,\mathrm{px}\,\mathrm{py}\,\mathrm{pz}\,\mathrm{r\_}\,x\,y\,z\right\}\right)$

$\left[\mathrm{quantumoperators}=\left\{\overrightarrow{L}\,\mathrm{Lx}\,\mathrm{Ly}\,\mathrm{Lz}\,\overrightarrow{p}\,\mathrm{px}\,\mathrm{py}\,\mathrm{pz}\,\overrightarrow{r}\,x\,y\,z\right\}\right]$

So for $\|L\|\^2$ and $L$ itself in terms of the vector operators $r$ and $p$, you have:

$\mathrm{LL}≔\mathrm{L\_}·\mathrm{L\_}$

$\mathrm{LL}≔\overrightarrow{L}·\overrightarrow{L}$

$\mathrm{L\_}≔\mathrm{r\_}\&x\mathrm{p\_}$

$\overrightarrow{L}≔\overrightarrow{r}\times \overrightarrow{p}$

where

$\mathrm{r\_}≔x\mathrm{\_i}+y\mathrm{\_j}+z\mathrm{\_k}$

$\overrightarrow{r}≔\stackrel{\wedge }{i}x+\stackrel{\wedge }{j}y+\stackrel{\wedge }{k}z$

$\mathrm{p\_}≔\mathrm{px}\mathrm{\_i}+\mathrm{py}\mathrm{\_j}+\mathrm{pz}\mathrm{\_k}$

$\overrightarrow{p}≔\stackrel{\wedge }{i}\mathrm{px}+\stackrel{\wedge }{j}\mathrm{py}+\stackrel{\wedge }{k}\mathrm{pz}$

The Commutator rules for the components $L_i$ are a consequence of the Commutator rules for the components of r and p. These rules can be set by using the Setup command, entering all of the commutators between any two of {$x,y,z,\mathrm{px},\mathrm{py},\mathrm{pz}$}, as in $\mathrm{Setup}\left(\mathrm{algebrarules}=\{\%\mathrm{Commutator}\left(x,\mathrm{px}\right)=I,\%\mathrm{Commutator}\left(x,\mathrm{py}\right)=0,\%\mathrm{Commutator}\left(x,y\right)=0,...\}\right)$. A convenient alternative for situations such as this is to create an indexing procedure for a Matrix. For example,

$C≔\left(a\,i\,b\,j\right)\mapsto \mathrm{\%Commutator}\left(\mathrm{Component}\left(a\,i\right)\,\mathrm{Component}\left(b\,j\right)\right)$

$C≔\left(a\,i\,b\,j\right)\mapsto {\left[{\left(a\right)}_i,{\left(b\right)}_j\right]}_{-}$

$\mathrm{algebra}≔\left(i\,j\right)\mapsto \left(C\left(\mathrm{r\_}\,i\,\mathrm{p\_}\,j\right)=\mathrm{I}\cdot \mathrm{KroneckerDelta}\left[i,j\right]\,C\left(\mathrm{r\_}\,i\,\mathrm{r\_}\,j\right)=0\,C\left(\mathrm{p\_}\,i\,\mathrm{p\_}\,j\right)=0\right)$

$\mathrm{algebra}≔\left(i\,j\right)\mapsto \left(C\left(\overrightarrow{r}\,i\,\overrightarrow{p}\,j\right)=\mathrm{I}{\mathrm{\delta }}_{i,j}\,C\left(\overrightarrow{r}\,i\,\overrightarrow{r}\,j\right)=0\,C\left(\overrightarrow{p}\,i\,\overrightarrow{p}\,j\right)=0\right)$

The commutators are then generated by the Matrix constructor, and the whole Matrix can be passed to Setup.

$\mathrm{Matrix}\left(3\,3\,\mathrm{algebra}\right)$

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

$\left[\mathrm{algebrarules}=\left\{{\left[\mathrm{px},\mathrm{py}\right]}_{-}=0\,{\left[\mathrm{px},\mathrm{pz}\right]}_{-}=0\,{\left[\mathrm{py},\mathrm{pz}\right]}_{-}=0\,{\left[x,\mathrm{px}\right]}_{-}=\mathrm{I}\,{\left[x,\mathrm{py}\right]}_{-}=0\,{\left[x,\mathrm{pz}\right]}_{-}=0\,{\left[x,y\right]}_{-}=0\,{\left[x,z\right]}_{-}=0\,{\left[y,\mathrm{px}\right]}_{-}=0\,{\left[y,\mathrm{py}\right]}_{-}=\mathrm{I}\,{\left[y,\mathrm{pz}\right]}_{-}=0\,{\left[y,z\right]}_{-}=0\,{\left[z,\mathrm{px}\right]}_{-}=0\,{\left[z,\mathrm{py}\right]}_{-}=0\,{\left[z,\mathrm{pz}\right]}_{-}=\mathrm{I}\right\}\right]$

The components of $L$ are:

$\mathrm{Lx}≔\mathrm{Component}\left(\mathrm{L\_}\,1\right)$

$\mathrm{Lx}≔y\mathrm{pz}-z\mathrm{py}$

$\mathrm{Ly}≔\mathrm{Component}\left(\mathrm{L\_}\,2\right)$

$\mathrm{Ly}≔z\mathrm{px}-x\mathrm{pz}$

$\mathrm{Lz}≔\mathrm{Component}\left(\mathrm{L\_}\,3\right)$

$\mathrm{Lz}≔x\mathrm{py}-y\mathrm{px}$

To verify that $\|L\|\^2=\mathrm{LL}$ commutes with each $L_i$,

$\mathrm{\%Commutator}\left(\mathrm{LL}\,\mathrm{Lx}\right)$

${\left[{\left(y\mathrm{pz}-z\mathrm{py}\right)}^2+{\left(z\mathrm{px}-x\mathrm{pz}\right)}^2+{\left(x\mathrm{py}-y\mathrm{px}\right)}^2,y\mathrm{pz}-z\mathrm{py}\right]}_{-}$

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

$0$

$\mathrm{Commutator}\left(\mathrm{LL}\,\mathrm{Ly}\right)$

$0$

$\mathrm{Commutator}\left(\mathrm{LL}\,\mathrm{Lz}\right)$

$0$

Depending on the case, an expansion of the Commutator is sometimes necessary to verify its value is 0, or to take the commutator rules into account in all subexpressions use Simplify.

2. Using a quantum-operator-tensor notation for the components of $L$, show that $\[L_i,L_j\]=I\mathrm{epsilon}\[i,j,k\]L_k$ (see the exercises of Chap VI in Cohen-Tannoudji). For this purpose, set the dimension of spacetime to 3, and its signature to Euclidean. To follow textbook notation, also use lowercaselatin letters for space tensor indices (see Setup).

$\mathrm{Setup}\left(\mathrm{dimension}=3\,\mathrm{signature}=\mathrm{`+`}\,\mathrm{spacetimeindices}=\mathrm{lowercaselatin\_is}\,\mathrm{spaceindices}=\mathrm{none}\,\mathrm{quiet}\right)$

$\left[\mathrm{dimension}=3\,\mathrm{signature}=\mathrm{+\; +\; +}\,\mathrm{spaceindices}=\mathrm{none}\,\mathrm{spacetimeindices}=\mathrm{lowercaselatin\_is}\right]$

Define $r$ and $p$ as tensors in this 3-D Euclidean space to be able to Simplify the result by using Einstein's summation convention for repeated (tensor) indices.

$\mathrm{Define}\left(r\,p\right)$

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

$\left\{p\,r\,{\mathrm{\gamma }}_i\,{\mathrm{\sigma }}_i\,{\partial }_i\,g_{i,j}\,{\mathrm{\epsilon }}_{i,j,k}\right\}$

Now set the related Commutator rules defining the algebra in tensor notation; in doing so, erase also previous settings (by using the redo option of Setup) for quantum operators and algebra rules (not necessary here, but sometimes desired).

$\mathrm{Setup}\left(\mathrm{redo}\,\mathrm{quantumoperators}=\left\{L\,p\,r\right\}\,\mathrm{algebrarules}=\left\{\mathrm{\%Commutator}\left(p\left[i\right]\,p\left[j\right]\right)=0\,\mathrm{\%Commutator}\left(r\left[i\right]\,p\left[j\right]\right)=\mathrm{I}\mathrm{kd\_}\left[i,j\right]\,\mathrm{\%Commutator}\left(r\left[i\right]\,r\left[j\right]\right)=0\right\}\right)$

$\left[\mathrm{algebrarules}=\left\{{\left[p_i,p_j\right]}_{-}=0\,{\left[r_i,p_j\right]}_{-}=\mathrm{I}{\mathrm{\delta }}_{i,j}\,{\left[r_i,r_j\right]}_{-}=0\right\}\,\mathrm{quantumoperators}=\left\{L\,p\,r\right\}\right]$

Verify how this algebra of Commutators works.

$\mathrm{\%Commutator}\left(r\left[j\right]\,p\left[k\right]\right)$

${\left[r_j,p_k\right]}_{-}$

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

$\mathrm{I}{\mathrm{\delta }}_{j,k}$

$\mathrm{Commutator}\left(r\left[j\right]p\left[k\right]\,r\left[m\right]p\left[m\right]\right)$

$-\mathrm{I}{\mathrm{\delta }}_{k,m}{r}_j{p}_m+\mathrm{I}{\mathrm{\delta }}_{j,m}{r}_m{p}_k$

Enter expressions for the tensor components of L that enter the commutator $\left[L_i\,L_j\right]$, to verify that it is equal to $I\mathrm{epsilon}\[i,j,k\]L_k$. In doing so, use the default abbreviation $\mathrm{\epsilon }$ for the LeviCivita pseudo-tensor.

$L\left[i\right]≔\mathrm{ep\_}\left[i,m,n\right]r\left[m\right]p\left[n\right]$

$L_i≔{\mathrm{\epsilon }}_{i,m,n}{r}_m{p}_n$

$L\left[j\right]≔\mathrm{subs}\left(i=j\,L\left[i\right]\right)$

$L_j≔{\mathrm{\epsilon }}_{j,m,n}{r}_m{p}_n$

Compute the Commutator.

$\mathrm{ans}≔\mathrm{Commutator}\left(L\left[i\right]\,L\left[j\right]\right)$

$\mathrm{ans}≔{\mathrm{\epsilon }}_{i,m,n}{\mathrm{\epsilon }}_{j,k,l}\left(-\mathrm{I}{\mathrm{\delta }}_{k,n}{r}_m{p}_l+\mathrm{I}{\mathrm{\delta }}_{l,m}{r}_k{p}_n\right)$

The above is indeed $L_k$, but is missing some simplification of the contracted indices of the LeviCivita and KroneckerDelta tensors, and the indices of the quantum operators $r,p$ involved in the noncommutative products.

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

$\mathrm{-I}\left(r_j{p}_i-r_i{p}_j\right)$