# Schmidt decomposition manages to write a pure state using just d terms

Suppose $$|\psi\rangle$$ $$\in \mathrm{H_A}\otimes\mathrm{H_B}$$ is a pure state and we can write a representation of $$|\psi\rangle$$ like $$|\psi\rangle = \sum_j |\alpha_j\rangle|\beta_j\rangle$$, where $$|\alpha_j\rangle$$ and $$|\beta_j\rangle$$ are un-normalized states for systems A and B, respectively. How can I prove that the numbers of terms in such representation are greater or equal to the terms in a Schmidt decomposition? My attempt is to begin using the partial trace but I don´t get anything.

Start by writing your state $$|\psi\rangle$$ in terms of the Schmidt decomposition $$|\psi\rangle=\sum_i\sqrt{p_i}|u_i\rangle|v_i\rangle$$ where $$|u_i\rangle$$ and $$|v_i\rangle$$ are orthonormal bases.

You want to evaluate the partial trace $$\text{Tr}_2|\psi\rangle\langle\psi|.$$ Remember that when you take the trace, you can select any orthonormal basis that you want. It's particularly convenient to select $$\{|v_i\rangle\}$$ in this case, $$\text{Tr}_2|\psi\rangle\langle\psi|=\sum_i(I\otimes\langle v_i|)|\psi\rangle\langle\psi|(I\otimes |v_i\rangle)=\sum_ip_i|u_i\rangle\langle u_i|$$ (I've deliberately skipped a step or 2 in that calculation, that you'd want to fill in for yourself.)

What does this tell you? The rank of the reduced density matrix is $$d$$, the number of non-zero Schmidt coefficients.

Now, assume you have a decomposition of $$|\psi\rangle$$ where the set $$\{|\alpha_j\rangle\}$$ comprising $$d' elements. The reduced density matrix is $$\sum_j|\alpha_j\rangle\langle\alpha_k|\langle\beta_k|\beta_j\rangle.$$ The dimension of this matrix cannot be larger that $$d'$$, and hence is not equal to $$d$$. Such a decomposition is impossible.

• I don´t understand why this decomposition is impossible if d' is not equal to d Mar 31, 2022 at 16:44
• Take a special case, with $d=3$. Now imagine that $d'=2$ and that $|\alpha_1\rangle$ and $|\alpha_2\rangle$ are in the span of $|0\rangle$ and $|1\rangle$) (they must always be in some two-dimensional span). If you write out the partial trace, it must be a $2\times 2$ matrix because the only non-zero components are $|0\rangle\langle 0$, $|1\rangle\langle 0$, $|0\rangle\langle 1$ and $|1\rangle\langle 1$. It has rank at most 2. So it certainly doesn't have rank $d=3$. Apr 1, 2022 at 6:28

The question is equivalent to asking why the rank of a matrix $$A$$ (i.e. the number of its nonzero singular values) equals the smallest number of terms in any decomposition of $$A$$ in unit rank factors.

To state this more precisely, observe that any matrix $$A$$ can be written as a sum of unit-rank terms. If $$A$$ is $$n\times n$$, then a trivial decomposition using $$n^2$$ terms is $$A = \sum_{ij=1}^n A_{ij} |i\rangle\!\langle j|.$$ The SVD tells you you can also always write $$A$$ using only $$\operatorname{rank}(A)\le n$$ terms: $$A = \sum_{k=1}^{\operatorname{rank}(A)} s_k |u_k\rangle\!\langle v_k|,\tag1$$ with $$\mathbb{R}\ni s_k\ge0$$ the singular values and $$\{|u_k\rangle\},\{|v_k\rangle\}$$ orthonormal systems. Now the question is: suppose we have some decomposition of the form $$A = \sum_{k=1}^m c_k |a_k\rangle\!\langle b_k|,\tag 2$$ for some $$c_k\in\mathbb{C}$$ and $$|a_k\rangle,|b_k\rangle$$. Can we conclude from this that $$m\ge \operatorname{rank}(A)$$?

To see that this is so, observe that (1) also implies that the support of $$A$$ (i.e. the orthogonal complement to its ker, i.e. the dimension of the space of vectors which are not sent to $$0$$ by $$A$$) has dimension $$\operatorname{rank}(A)$$. This is because its support is comprised of all and only the vectors writable as linear combinations of $$|v_k\rangle$$ with $$k=1,...,\operatorname{rank}(A)$$.

Now, suppose we have (2) with $$m<\operatorname{rank}(A)$$. This means that the support of $$A$$ is comprised of all and only the vectors in the span of $$m$$ vectors. This space is bound to have dimensional smaller than $$\operatorname{rank}(A)$$. Which means there are up to $$\operatorname{rank}(A)$$ orthogonal vectors not in the ker of $$A$$. But that's a contradiction, because from the SVD we now that there are $$\operatorname{rank}(A)$$ orthogonal vectors not in the ker of $$A$$.