Formal problem statement. For a real vector $\vec r$ in bases of $\sigma_i\otimes \sigma_j$ with $i,j=0,1,2,3$, $i$ and $j$ not equal to $0$ at the same time, $\sigma_0=I$ and other indexes stands for Pauli matrices. There are $15$ of them. How to prove that under the action of unitary operation(4 by 4 matrix), i.e. $U\sum_{ij}r_{ij}\sigma_i\otimes \sigma_j U^\dagger$, the length(Euclidean norm) of $\vec r$ remain unchanged(in the prescribed bases)? We can only consider SU(4) elements.

Here are some of my thoughts. I refer to the 'qubit' case for some inspiration, i.e., the bases are only $3$ Pauli matrices, and the vector has $3$ dimensions. A brute-force proof is to check the square of the coefficients of $U\sigma_1U^\dagger$ sum to 1, the square of the coefficients of $U\sigma_2U^\dagger$ sum to 1, and also the square of coefficients of $U\sigma_3U^\dagger$ sum to 1. Then for any term $\sum_{i=1,2,3}r_i\sigma_i$, after the action of unitary, i.e. $U\sum_{i=1,2,3}r_i\sigma_i U^\dagger$, easy to see the length of $\vec r$ remain unchanged. But the method becomes too complicated in the '$2$-qubit' case for we need to check $15$ matrix equations in total. So for the '$2$-qubits' case, is there some easier method, and also, for the '$d$-qubit' case, is this character still remain?


2 Answers 2


One of the key things that does not change as the result of a unitary transformation applied to a matrix is the eigenvalues of the matrix. Consequently, any function of the matrix that only depends on the eigenvalues and not the eigenvectors also does not change. If we can see that the length of $\vec{r}$ is one of these, we're done.

I like your thought about using the one-qubit case for inspiration. That's what helped me. In the 1-qubit case, $M=\sum_{i\in\{1,2,3\}}r_i\sigma_i$ has eigenvalues $\pm\sqrt{r_1^2+r_2^2+r_3^2}$. A way to prove this is to consider $\text{Tr}(M^2)=\sum_i\lambda_i^2$ ($\lambda_i$ are the eigenvalues), and realising that by omitting the $I$ term, you have $\sum_i\lambda_i=0$.

If you do the same thing in the two-qubit case, you'll get $$ \text{Tr}(M^2)=\sum_i\lambda_i^2 $$ which, being only a function of eigenvalues, is invariant under the action of $U$.

So, now evaluate $$ \text{Tr}(M^2)=4\sum_{i,j}r_{ij}^2=4\|\vec{r}\|^2. $$ (This works, because by taking the trace of the product, you only pick out $I$ terms, which only arise when a tensor product of Paulis multiplies itself, and has trace 4.)


Observe the following:

  1. When representing a state in the Pauli basis, or more generally, when representing a state by decomposing it in a basis of orthogonal Hermitian operators that includes the identity (often referred to as a "Bloch representation"), the length of the corresponding vector of coefficients is related to the purity of the represented state. This is because if $$\rho = \frac{1}{D}(I + \sum_k c_k \sigma_k),\qquad \mathrm{Tr}(\sigma_k)=0, \, \mathrm{Tr}(\sigma_i\sigma_j)=D\delta_{ij},$$ then $\mathrm{Tr}(\rho^2)=\frac{1}{D}(1+\|\vec c\|^2).$

  2. The purity of a state depends exclusively on its eigenvalues: $\mathrm{Tr}(\rho^2)=\sum_k p_k^2$ with $p_k$ eigenvalues of $\rho$. You can see this as a special case of the more general fact that $\mathrm{Tr}(A^\dagger A)$ equals the sum of the squares of the singular values of $A$.

  3. Unitary evolution does not change the eigenvalues of a state: $U\rho U^\dagger$ always has the same spectrum as $\rho$.

It follows that unitary operations preserve the length of the vector of coefficients representing the state in the Bloch representation/Pauli basis. This is true for states in any dimension. Another way to frame this is to notice that unitary matrices act on density matrices via their adjoint representation as rotation matrices. This is discussed in Can we rotate Bloch vectors for qudits like we do with qubits in the Bloch sphere?.


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