Matrix Representation of Tensor Products

1. The Action of a Combined Operator

Before we get to the matrix calculation, let's establish the intuition. Suppose you have two vector spaces, $V$ and $W$, and two linear operators, $A$ acting on $V$ and $B$ acting on $W$. We want to define a new operator, which we'll call $A\otimes B$, that acts on the combined space $V\otimes W$.

The most natural way to define this is to say that the combined operator acts on each part of a simple tensor product state separately. That is, for any vectors $\mathbf{v}\in V$ and $\mathbf{w}\in W$, the action is defined as:

This rule is the conceptual core of everything that follows. It says: "To apply the combined operator $A\otimes B$, just apply $A$ to the first part of the state and $B$ to the second part, then take the tensor product of the results."

The following video discusses this principle from the perspective of quantum mechanics, where it's used constantly.

2. Constructing the Matrix: The Kronecker Product Revisited

In the last lesson, we used the Kronecker product to build a larger vector from two smaller vectors. It turns out the exact same procedure works for matrices. This provides the concrete recipe for constructing the matrix of the operator $A\otimes B$.

The rule is: if $A$ is an $m\times n$ matrix and $B$ is a $p\times q$ matrix, their Kronecker product $A\otimes B$ is an $(mp)\times (nq)$ block matrix. To build it, you multiply each scalar entry of $A$ by the entire matrix $B$.

Let's watch a video that defines and demonstrates this process.

The image below, which you may recognize from the video, provides a great static visual of this calculation.

3. Connecting the Matrix and the Action

Now for the crucial connection. The matrix we just learned to construct using the Kronecker product is precisely the matrix representation of the operator defined by the action (A ⊗ B)(v ⊗ w) = (Av) ⊗ (Bw).

This is a fundamental property that makes the Kronecker product the "right" way to combine matrices for composite systems. It ensures that our computational method (building the big matrix) correctly implements our conceptual goal (acting on each part of the system separately).

The Michael Penn video you just watched concludes by explicitly stating this key property.

4. Application: Applying a Quantum Gate to a Single Qubit

Let's make this concrete with an example central to your goal of learning quantum mechanics.

Consider a two-qubit system. How would we apply an operation to the first qubit, but leave the second qubit unchanged? The operator for "do nothing" is the identity matrix, $I$. If the operation on the first qubit is, say, the bit-flip gate $X$ (the quantum equivalent of a NOT gate), the combined operator is $X\otimes I$.

The matrices for these single-qubit operators are:

Let's construct the matrix for $X\otimes I$:

Now, let's apply this 4x4 matrix to a two-qubit state, say $|01⟩$, which means the first qubit is in state $|0⟩$ and the second is in state $|1⟩$. In the last lesson, we found that $|01⟩=|0⟩\otimes |1⟩=\left(\begin{array}{c}1\\ 0\end{array}\right)\otimes \left(\begin{array}{c}0\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right)$.

Let's perform the matrix-vector multiplication:

Wait, something is wrong here. Let me re-calculate.
$|0⟩=[1,0{]}^T,|1⟩=[0,1{]}^T$.
$|01⟩=|0⟩\otimes |1⟩=[1,0{]}^T\otimes [0,1{]}^T=[1\cdot [0,1{]}^T,0\cdot [0,1{]}^T{]}^T=[0,1,0,0{]}^T$. Correct.

Let's do the matrix multiplication again:

Ah, that's better. The resulting vector is $(0,0,0,1{)}^T$. What state is this? This is exactly the vector for $|11⟩=|1⟩\otimes |1⟩$.

Let's check this using our conceptual rule:

We know that $X|0⟩=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)=|1⟩$, and $I|1⟩=|1⟩$.
So the result is $|1⟩\otimes |1⟩=|11⟩$. It works perfectly! The matrix we constructed performs exactly the operation we intended.

This article provides a brief written walkthrough of this exact example, which will help solidify your understanding.

Conclusion

Today we bridged the gap between states and the operators that act on them in composite systems. By extending the Kronecker product from vectors to matrices, we found a concrete method for building the matrix representation of a combined operator.

Key Takeaways:

• The tensor product of two operators, $A\otimes B$, is defined by its action on a simple product state: $(A\otimes B)(\mathbf{v}\otimes \mathbf{w})=(A\mathbf{v})\otimes (B\mathbf{w})$.
• The matrix for this operator is constructed using the Kronecker product of the individual matrices, $A$ and $B$, following a "scale-and-place" block matrix rule.
• The dimensions of the resulting matrix are the products of the original dimensions: if $A$ is $m\times n$ and $B$ is $p\times q$, then $A\otimes B$ is $(mp)\times (nq)$.
• This construction allows us to model operations that act on specific parts of a composite system, like applying a quantum gate to a single qubit by using an operator like $G\otimes I$.

Preview of the Next Lesson:

So far, we have only worked with "simple" tensor product states of the form $\mathbf{v}\otimes \mathbf{w}$. But the tensor product space $V\otimes W$ contains many more vectors—namely, linear combinations of these simple products (e.g., ${\mathbf{v}}_1\otimes {\mathbf{w}}_1+{\mathbf{v}}_2\otimes {\mathbf{w}}_2$). In the next lesson, we will explore the crucial distinction between states that can be written as a simple product (separable states) and those that cannot (entangled states). This is one of the most fascinating and counter-intuitive features of quantum mechanics, and its foundation lies in the linear algebra we've just covered.