3
$\begingroup$

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.

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

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 $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.