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.


2 Answers 2


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 $d'<d$ 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.

  • $\begingroup$ I don´t understand why this decomposition is impossible if d' is not equal to d $\endgroup$
    – username9
    Mar 31, 2022 at 16:44
  • $\begingroup$ 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 $d=3$. $\endgroup$
    – DaftWullie
    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 $A$.


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