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According to schmidt decomposition if I have pure state $|\psi\rangle$ in the composite hilbert space $AB$ ( both $A$ and $B$ are hilbert spaces of dimension $n$ ) then it can be writen as $$|\psi\rangle = \sum_i \lambda_i |i_A\rangle |i_B\rangle$$ here $\{i_A\}$ and $\{i_B\}$ are orthonormal basis for hilbert spaces $A$ and $B$ respectively.
The above holds true for any vector $|\psi\rangle$ ( if I am not wrong ) even if its not normalized ( if not normalized then $\sum_i \lambda_i^2$ wont be equal to 1 ). Thus any vector of of a space of dimension $n$ x $n$ is being written in linear combination of only $n$ orthonormal vectors ( $|i_A\rangle |i_B\rangle$ for $1 <= i <= n$ ), which should not be possible. Am I missing something or interpreting schmidt decomposition incorrectly ?

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    $\begingroup$ It is not always the same couple of bases. The theorem says that given a vector $\psi$, it is possible to choose two bases (that depend on $\psi$) such that the result holds. $\endgroup$ – yuggib Apr 28 '15 at 21:34
  • $\begingroup$ @yuggib oh I get it now, as pair of basis might differ they don't belong to a single n dimensional space. am I correct ? $\endgroup$ – sashas Apr 28 '15 at 21:48
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    $\begingroup$ The idea is that you cannot write any $\psi$ as a different combination of the same basis, but for any $\psi$ you can choose a basis that do the job. So the dimensionality of the whole space is still $n\times n$, and it is not spanned by an $n$ dimensional basis. $\endgroup$ – yuggib Apr 28 '15 at 22:14
  • $\begingroup$ @yuggib Although not a full answer, I really think these are worthwhile recording in a permanent answer as this is a key point that may be helpful to others. $\endgroup$ – Selene Routley Apr 29 '15 at 6:43
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Denote $|\psi\rangle = \sum\limits_{i = 1}^m \sum\limits_{j = 1}^n h_{ij} |ij\rangle$ as $|\psi\rangle \rightarrow H = (h_{ij})_{m \times n}$. Then we have the following lemma:

Lemma: Define matrix $U$ (in the original basis) as a new setting for Alice and $V$ for Bob, then a state $|\psi\rangle$ in the basis of the new settings is $U^* H V^\dagger$.

Proof: Denote the original bases of Alice and Bob both as $|0\rangle, |1\rangle$, the new setting as $|0_a\rangle, |1_a\rangle$ and $|0_b\rangle, |1_b\rangle$ for Alice and Bob, respectively. and so $$ \begin{bmatrix} |0_a\rangle\\ |1_a\rangle \end{bmatrix} = U \begin{bmatrix} |0\rangle\\ |1\rangle \end{bmatrix}, \begin{bmatrix} |0_b\rangle\\ |1_b\rangle \end{bmatrix} = V \begin{bmatrix} |0\rangle\\ |1\rangle \end{bmatrix}. $$ In this way, the state is expressed as $$ |\psi\rangle = [|0\rangle, |1\rangle] H \begin{bmatrix} |0\rangle\\ |1\rangle \end{bmatrix} = [|0_a\rangle, |1_a\rangle] (U^\dagger)^T H V^\dagger\begin{bmatrix} |0_b\rangle\\ |1_b\rangle \end{bmatrix}, $$ Thus, in the basis of new setting, the state is expressed as $U^* H V^\dagger$.

Linear algebra tells us that: Given a matrix $H$, there are some unitary matrices $U,V$ such that $U^* H V^\dagger$ is a diagonal matrix. So, given any state $|\psi\rangle$, there are some basses $|0_a\rangle, |0_b\rangle$ such that $|\psi\rangle$ can be expressed as a diagonal matrix ${\rm Diag}(\lambda_1,\lambda_2,\cdots,\lambda_n)$, that is, $$ |\psi\rangle = \sum_i \lambda_i |0_a\rangle |0_b\rangle. $$
Obviously, $|0_a\rangle, |0_b\rangle$ may be different for different $|\psi\rangle$.

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As suggested by WSA aka RV, I copy my comments into a (partial) answer.

The key point is that the theorem says "for any given $\psi$ there exist two bases $\{i_A\}$ and $\{i_B\}$ such that...". This means that the choice of the bases depends on the vector $\psi$ we are considering.

So there is not an $n$ dimensional common basis that span the whole $n\times n$ space; and if we would be very precise we may write the formula as $$\psi= \sum_{i=1}^n \lambda_i(\psi)i_A(\psi)\otimes i_B(\psi) \; ,$$ i.e. clarifying the $\psi$-dependence of the bases (and coefficients obviously).

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