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Here I have a two pure state system composed from systems A and B: $$\Psi_{ab} = \sqrt{p}\, |0\rangle_a |0\rangle_b + \sqrt{1-p}\, |1\rangle_a |1\rangle_b $$

How do I extract system A in matrix form using trace?

My attempt: $$ \hat A = \mathrm{Tr}_b\,( |\Psi_{ab} \rangle \langle \Psi_{ab}|) = \sum_N \langle n_b| \Psi_{ab} \rangle \langle \Psi_{ab}| n_b \rangle $$

which makes

$$ \hat A = \left(\begin{matrix} p & 0 \\ 0 & 1-p \end{matrix}\right) $$

however, my confusion is that if I were to look for the operator of B then I would also get the same answer. Is my understanding of Trace for a Hilbert space (A or B) incorrect?

Any help would be most appreciated. Thanks.

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    $\begingroup$ Since your system is unchanged under $a\leftrightarrow b$, isn’t it expected that $Tr_a=Tr_b$? $\endgroup$ Apr 22 at 3:53
  • $\begingroup$ @ZeroTheHero I think so, my understanding of this is just fairly shaky. But that is my initial intuition, I am just not sure if that is the correct one. $\endgroup$ Apr 22 at 5:50
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Let me illustrate this from a more general perspective: For a pure state of a bipartite system, the corresponding reduced density matrices have the same non-zero eigenvalues.

To see this, consider a Hilbert space $\mathscr{H} = \mathscr{H}_1 \otimes \mathscr{H}_2$ and let $\{|a_i\rangle\}_i$ and $\{|b_j\rangle\}_j $ denote complete and orthonormal bases sets in $\mathscr{H}_1$ and $\mathscr{H}_2$, respectively. Then a generic state $|\psi\rangle \in \mathscr{H}$ can be expanded as follows:

$$ |\psi\rangle = \sum\limits_{ij}c_{ij}\,|a_i\rangle \otimes |b_j\rangle \quad .$$

However, using the Schmidt decomposition, we can also write this state as

$$|\psi\rangle = \sum\limits_{\alpha} \sigma_{\alpha} \,|A_\alpha\rangle \otimes |B_\alpha\rangle \quad, $$

with $\sigma_\alpha >0$, $\sum\limits_{\alpha} \sigma_{\alpha}^2 = 1$ and $\{|A_{\alpha}\rangle\}_{\alpha}$ as well as $\{|B_{\alpha}\rangle\}_{\alpha}$ denoting complete orthonormal sets in $\mathscr{H}_1$ and $\mathscr{H}_2$, respectively.

Defining the density operator $\rho_{\psi} \equiv |\psi\rangle\langle\psi|$, we can compute its reduced density matrices $\rho_1\equiv \mathrm{Tr}_2 \,\rho_{\psi}$ and $\rho_2\equiv \mathrm{Tr}_1\,\rho_{\psi}$ in a straightforward manner and find that they have the same non-zero eigenvalues, namely $\sigma_{\alpha}^2$:

\begin{align} \rho_1 &= \sum\limits_\alpha \sigma_{\alpha}^2 \, |A_{\alpha}\rangle \langle A_{\alpha}| \\ \rho_2 &= \sum\limits_\alpha \sigma_{\alpha}^2 \, |B_{\alpha}\rangle \langle B_{\alpha}| \quad . \end{align}


In conclusion, your result perfectly fits in this scheme, as the wave function you have given is already in its Schmidt decomposition.

A more detailed treatment of this should be given in any textbook on quantum information theory.

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  • $\begingroup$ Thank you! I didn't even know that this fits into quantum information theory, this is material from my quantum computing class. I now know where to look in the future. $\endgroup$ Apr 23 at 1:28

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