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The total angular momentum operator isUsing the cyclic orbital angular momentum commutation relations,while $[L_z,L_z^2]=0$; hence $\boxed{[L_z,{\mathbf{L}}^2]=0}$. In Cartesian coordinates,Because $L_z(x+iy)=\hslash (x+iy)$ and it annihilates $z$ and every radial function,Replacing $i$ by $-i$ gives eigenvalue $-m\hslash$.

The six-dimensional space splits into the five trace-free quadratic spherical harmonics with $\ell =2$ and one radial state with $\ell =0$. A simultaneous eigenbasis for $L_z$ and ${\mathbf{L}}^2$ iswhere the eigenvalue $\ell (\ell +1){\hslash }^2$ of ${\mathbf{L}}^2$ was used. Every displayed state is a complex linear combination of the original six energy eigenstates, so it is still an energy eigenstate at the same level.

Let $r=dim(U\cap W)$. Choose a basis $e_1,\dots ,e_r$ of the intersection of vector subspaces, extend it to a basis $e_1,\dots ,e_r,u_{r+1},\dots ,u_m$ of $U$, and independently extend it to a basis $e_1,\dots ,e_r,w_{r+1},\dots ,w_n$ of $W$. The combined listspans the sum of vector subspaces $U+W$. It is also linearly independent: a vanishing linear combination says that a combination of the $u_i$ belongs to $U\cap W$, and its expression in the chosen basis of $U$ forces all its coefficients to vanish; the coefficients of the $w_i$ then vanish in the same way. This proves the dimension formula for a sum of subspaces

Parametrize the two linear subspaces asPutting a vector of $U$ in $W$ first forces its fourth coordinate to vanish; the remaining defining equation is then automatic. Thusand the dimension formula gives $\dim (U+W)=2+2-1=3$.

The linear functionalvanishes on each displayed generator of $U$ and $W$. Its kernel is three-dimensional because $\ell \ne 0$, so the inclusion $U+W\subseteq \ker \ell$ between two three-dimensional subspaces is equality.

Using the states from part (a),The Born rule therefore givesAfter the equal-sign outcomes the state is respectively $|\uparrow \uparrow ⟩$ or $|\downarrow \downarrow ⟩$. These are triplet states, hence eigenstates of ${\mathbf{S}}^2$ with eigenvalueThe opposite-sign product states are each a nontrivial linear combination of the singlet and the $m=0$ triplet, so they are not eigenstates of ${\mathbf{S}}^2$.

A normalized quantum state is a linear combination of orthonormal basis vectors in a Hilbert space. Its density operator is $|\psi ⟩⟨\psi |$. When at least two basis amplitudes are nonzero, it has off-diagonal terms and differs from the mixed state ${\sum }_j|a_j{|}^2|j⟩⟨j|$. Quantum gates act linearly on this superposition and preserve its phase information.

In a chain $N_1\to N_2\to N_3$ with distinct decay constants, each population is a linear combination of the exponentials $e^{-{\lambda }_{i}t}$. When two consecutive constants coalesce to $\lambda$, the repeated Laplace-transform pole produces a term proportional to $t{e}^{-\lambda t}$.

If ${\sum }_i{g}_i{f}_i=1$, expand its power $r(N-1)+1$. By the pigeonhole principle, each monomial contains some $f_i^N$. Collecting terms expresses $1$ as a linear combination of these powers. This lets one use a common denominator exponent in localization arguments.