# Is the partial trace of a Schmidt decomposition of a pure state a spectral decomposition?

I have a doubt about the spectral decomposition of the partial trace of the Schmidt decomposition.

• $$\mathcal{H}_a$$ and $$\mathcal{H}_b$$ are two Hilbert spaces with $$\text{dim}(\mathcal{H}_a) = N$$ and $$\text{dim}(\mathcal{H}_b) = M$$
• $$|u_k\rangle$$ is a set of orthonormal vectors of $$\mathcal{H}_a$$
• $$|v_k\rangle$$ is a set of orthonormal vectors of $$\mathcal{H}_b$$
• $$d = \min(N, M)$$
• $$|\phi\rangle \in \mathcal{H}_a \otimes \mathcal{H}_b$$

Now the Schmidt decomposition of $$|\phi\rangle$$ is: $$|\phi\rangle = \sum_{k=0}^{d-1} \lambda_k |u_k\rangle |v_k\rangle \quad .$$

Its partial trace on $$\mathcal{H}_b$$ is:

$$\phi_a = \sum_{k=0}^{d-1} \lambda_k^2 |u_k\rangle \langle u_k|\quad .$$

Is the result of the partial trace of the Schmidt decomposition a spectral decomposition? And if it is a spectral decomposition, how could it be? Because $$|u_k\rangle$$ is a set of orthonormal vector but not a basis of $$\mathcal{H}_a$$ and $$\phi_a$$ is an operator on $$\mathcal{H}_a$$ but the eigenvectors $$|u_k\rangle$$ don't form a basis for $$\mathcal{H}_a$$.

• Your last equation makes no sense. On the LHS there is a vector, on the RHS an operator. Despite that, I don't understand the question. You can extend the orthonormal system to an orthonormal basis...so yes, it is a spectral decomposition. Commented May 4 at 19:51
• Your edit did not fix the first point of my comment...$\phi_a$ is not an element of $H_a$; it is an operator on it. Commented May 4 at 19:57
• I've correct, sorry I'm slow to write Commented May 4 at 20:07
• My doubt is that if the operator is hermitian the eigenvectors of the spectral decomposition are a base for the space but the uk are only a set of orthonormal vectors and not a base for the space Commented May 4 at 20:09
• To have a basis you only have to have the right number of vectors and they must be linearly independent, they do not even have to be orthonormal vectors. But even if they are not a basis (maybe there are not enough...) they can still give the spectral decomposition of an operator if that operator has a low rank. you then only need a partial basis. Commented May 4 at 20:20

This question lacks some core information about Schmidt decomposition. Let me clarify it here:

If we consider a state vector $$|\psi \rangle$$ in a composite Hilbert space $$H_A \otimes H_B$$ by saying that $$|\alpha\rangle$$ forms a basis in $$H_A$$ and $$|\beta\rangle$$ forms a basis in $$H_B$$ with dimensions $$n$$ and $$m$$, then it can be be represented as: $$|\psi\rangle = \sum_{\alpha \beta} C_{\alpha \beta} |\alpha\rangle \otimes |\beta\rangle$$

Now, if we assume that $$d = min{\alpha,\beta}$$ and the rank of matrix $$C$$ is $$k$$, then it can be safely assumed that $$k \leq d$$. Using the SVD, the matrix C can be written as $$C = U \Sigma V$$, with $$U$$ and $$V$$ as unitary operators with dimensions $$n \times n$$ and $$m \times m$$. $$\Sigma$$ is a $$n \times m$$ matrix such that $$\Sigma_{i,j} = 0 \text{if} i \neq j$$ or $$i = j = s> k$$. Using the above information, the state $$|\psi \rangle$$ can be expressed in the following way:

$$|\psi\rangle = \sum_{i,j} \Sigma_{i,j} (\sum_\alpha U_{\alpha,i} |\alpha\rangle) \otimes (\sum_{\beta} V_{j,\beta} |\beta\rangle) = \sum_{i=0}^k \Sigma_{i,i} |\alpha_i\rangle \otimes |\beta_i\rangle$$ Where $$|\alpha_i\rangle = \sum_\alpha U_{\alpha,i} |\alpha\rangle$$ and $$|\beta_i\rangle = \sum_{\beta} V_{i,\beta} |\beta\rangle$$. The following points should be noted now:

1. The set $$|\alpha_i\rangle$$ and $$|\beta_i\rangle$$ form a basis in Hilbert spaces $$H_A$$ and $$H_B$$.
2. Now, if we take the partial trace of $$\rho = |\psi\rangle \langle \psi|$$ over B, then we get $$\rho_A = \sum_i \Sigma_{i, i}^2 |\alpha_i\rangle \langle \alpha_i| \implies$$ that the decomposition is a spectral decomposition.
• thanks, so I've $n$ $|a_i\rangle$ but in the schmdit decomposition I use only $k$ of these vectors? Commented May 6 at 18:55
• So, if $\rho_A$ is a low-rank operator, it can be described in the spectral decomposition by a number of eigenvectors fewer than that of the space in which it operates? Commented May 6 at 19:20
• Yes. In other words, we can say that the other eigenvectors correspond to zero eigenvalues, so they are also always present there, even if the reduced state has a lower rank. Commented May 6 at 19:59