# Tensor product of operators in QM

If I wanted to find the coefficients of a linear transformation between 2 vectors in the basis for 2 spin $1/2$ paticles (let's say for starters we are not even looking for a unitary transform):

$$Z=\{|++\rangle,|+-\rangle,|-+\rangle,|--\rangle\},$$

should I look for 16 coefficients,

or should I use $A\otimes B$, with $A$ and $B$ $\in L(\mathbb{C}^2)$, which would be only 8 coefficients, but with a structure imposed by the structure of the Kronecker product:

$$A\otimes B = \left(\begin{array}{cc} a_{11} B & a_{12} B \\ a_{21} B & a_{22} B \end{array}\right)$$

or should I use (edit: this is wrong, read Lubos answer below):

$$A\otimes B=A\otimes \mathbb{1}+\mathbb{1}\otimes B,$$

in which case there is again only 8 coefficients to find, but the structure seems to be different. Note that this last form is what we use when figuring out the total Hamiltonian for a system of 2 particles.

I am very confused about this - actually - I'm guessing that transformations that can be written $A\otimes B$ are a subset of all possible transformations with all 16 coefficients free. But I'm still confused about the last 2 expressions above: if I had to apply the operator "$A\otimes B$" to e.g, $|+-\rangle$, I would not be confident about which expression to use for $A\otimes B$.

First of all, the equation $$$$A\otimes B=A\otimes \mathbb{1}+\mathbb{1}\otimes B,$$$$ is a claim about an identity, and this claim is incorrect. Note that for $1\times 1$ matrices, the matrices are numbers and the equation above reduces to $$a\cdot b = a\cdot 1 + 1 \cdot b$$ which is clearly wrong because the addition (the right hand side is just $a+b$) is something else than multiplication!

If we have a Hilbert space $H = H_A \otimes H_B$ and there is a change of the basis in the Hilbert space $H_A$ separately and in $H_B$ separately, the change of the basis in the Hilbert space $H = H_A \otimes H_B$ is the tensor product of the two transformation matrices, $A\otimes B$, in your notation (also expressed in the block form using those $a_{11}B$ etc.), and not $A\otimes 1 + 1\otimes B$.

This is easily calculated. The basis vectors transform as $$a_j = a'_i \cdot A_{ij}, \quad b_j = b'_i\cdot B_{ij}$$ (or switch the order of the indices or which side has the prime or invert the matrix or transpose it etc. – at least some of these choices are conventions; you must be careful to use consistent conventions all the time) and the basis vectors of $H_A\otimes H_B$ are $$a_j\otimes b_\ell = a'_i \cdot A_{ij}\otimes b'_k\cdot B_{k\ell} = (a'_i\otimes b'_k) \cdot A_{ij}B_{k\ell}$$ The rule here is that algebraically, the tensor multiplication is treated just like a normal multiplication (with no identification of indices or summing over indices). The transformation matrix in $H_A\otimes H_B$ is therefore $A_{ij}B_{k\ell}$ where $i,k$ are parts of the "first" index generalizing $i$ or $k$, and $j,\ell$ are analogously the second index.

The expression $$A\otimes 1 + 1\otimes B$$ is also important and it appears at many places, but we must discuss the "derivatives", the changes under the infinitesimal transformations. It's because the expression above is analogous to the Leibniz rule for the derivative of the product $$(ab)' = a'b + ab'$$ In quantum mechanics, derivatives may also be represented as commutators. For example, if $\vec J$ is the operator of the total angular momentum, then $$[J_z,A]$$ is the operator measuring the $z$-component of the angular momentum carried by the operator $A$. Similarly for $B$. Because we may decompose the commutator with a product to the sum of two commutator-like terms, $$[J_z,AB] = J_z AB - AB J_z = J_z AB - A J_z B + A J_z B - AB J_z = [J_z,A]B+A[J_z,B]$$ one may see that the result of acting with $J_z$ on the whole $AB$ is a sum of two terms – the angular momentum of $A$ and the angular momentum of $B$. Those give the two terms you mentioned because the angular momentum of $A$ only acts as $J_z^A$ on the space $H_A$, but it acts as the identity operator (one) on the space $H_B$, and vice versa, and that is why you have the tensor products with the identity matrix.

So the expressions $A\otimes 1+1\otimes B$ would appear but not if you considered a "finite" change of the bases, but if you only considered changes of bases that are "infinitesimally close" to the identity change (no change at all), and if you would take a difference between these two expressions (a derivative with respect to some parameters labeling the change of the bases, taken near the trivial transformation).

In terms of groups and algebras, the straightforward $A\otimes B$ gives us the matrix of a finite transformation in the Lie group as it acts on the tensor-product representation $H_A\otimes H_B$. On the other hand, $A\otimes 1+1\otimes B$ is the form of the Lie algebra generator with respect to the same space. Elements of Lie algebras are matrices/operators $(g-1)/\epsilon$ where $g$ is a group element infinitesimally close to the identity $1$ of the Lie group.

• Thanks Lubos - this is an invaluable explanation for me. I added an "edit" to my initial post to make sure people see right away that $A\otimes B \neq A\otimes\mathbb{1}+\mathbb{1}\otimes B$ in general. – Frank Apr 30 '15 at 13:50
• One more question - I have seen $H_A\otimes\mathbb{1}+\mathbb{1}\otimes H_B$ for the total Hamiltonian of a system of two spin 1/2 particles. Can that expression be related to the derivative of something? (this is now a specific QM question about the Hamiltonian for systems of 2 particles) – Frank Apr 30 '15 at 14:11
• Dear Frank, thanks for the "non equal" fix! And concerning the question: absolutely. This is an example I should have written. The Hamiltonian is effectively the "derivative of the state vector with respect to time" over $i\hbar$, as Schrödinger's equation says. If the wave function is $\psi_A\otimes \psi_B$, a tensor product of two independent subsystems, then the time derivative acts on this product via the Leibniz rule, $(\psi_A\otimes \psi_B)' = \psi'_A\otimes \psi_B+\psi_A\otimes \psi'_B$ and this can also be written as $(1/i\hbar)(H_A\otimes 1+1\otimes H_B)$ acting on $\psi$ – Luboš Motl Apr 30 '15 at 17:10
• because e.g. $\psi'_A\otimes \psi_B = (1/i\hbar) H_A\psi_A \otimes \psi_B = (1/i\hbar) (H_A\otimes 1) (\psi_A\otimes \psi_B)$. So the Hamiltonian is an example of that "derivative" because it's a time derivative.Similarly, the momentum is the derivative with respect to space, angular momentum is the derivative with respect to the angles (rotations), like in Noether's theorem. – Luboš Motl Apr 30 '15 at 17:12
• Is it true also if the systems interact? Probably not, right? Where do you insert interaction terms if the two systems interact? – Frank Apr 30 '15 at 17:41

I am very confused about this - actually - I'm guessing that transformations that can be written $A⊗B$ are a subset of all possible transformations with all 16 coefficients free.
This is correct. First some notation: Let $H_1$ and $H_2$ be two (finite-dimensional) Hilbert spaces for two quantum systems, and $End(H_1), End(H_2)$ be the vector space of operators acting on these spaces. As you are aware if we want to consider both systems at the same time, we consider the tensor product $H_1 \otimes H_2$. Let $dim(H_1) = n$ with basis $\{v_i\}$ and $dim(H_2) = n$ with basis $\{w_j\}$, and note that since $H_1 \otimes H_2$ has basis $\{v_i \otimes w_j\}$, $dim(H_1 \otimes H_2) = nm$. This means that the most general vector $x \in H_1 \otimes H_2$ can be written as $$x = \sum_{i,j} c_{ij} v_i \otimes w_j ,$$ so we need $nm$ coefficients to specify a vector here. There is also the subspace which consists of "pure" states, which can be factored as $x = v \otimes w$. This is a subset of all possible vectors, and is in fact isomorphic to $H_1 \times H_2$, the direct product (also commonly written as $H_1 \oplus H_2$, and called the direct sum). In other words, the tensor product contains a copy of the direct product, which has dimension $n+m$, so NOT every state can be written in a factored way, since $n+m < nm$ (assuming dimensions of these spaces is greater than one).
The situation for operators acting on $H_1 \otimes H_2$ is exactly similar: Naturally, we have $End(H_1 \otimes H_2) = End(H_1)\otimes End(H_2)$, which would have dimension $n^2m^2$ (=16 for your example), and contains a subspace isomorphic to $End(H_1) \times End(H_2)$ which consists of operators of the form $A\otimes B$, which has dimension $n^2 + m^2$ (= 8 for your example).