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During my studies of composite quantum systems I find some expressions that leave me with a little doubt. For example: Let K be a linear operator defined in the Hilbert space H. Where H is given by $H = H_{a}\otimes H_{b}$. If I want to perform the trace of $K$ in $H$ space, I use the following expression $$ \mathrm{Tr}(K) = \sum_{j=1}^{n}\langle\psi_j|K|\psi_j\rangle $$ where {$|\psi_j\rangle$} is some basis on $H$. With that, if i want to perform the partial trace of this operator over some basis of $H_b$ given by {$|b_j\rangle$}, i use the following expression:

$$ \mathrm{Tr}_b(K) = \sum_{j}(I_a\otimes\langle b_j|)K(I_a\otimes|b_j\rangle)\tag{1} $$
I have doubts in this expression:

Is $I_a$ is the identity operator given by $\sum_{i}|a_i\rangle\langle a_i|$?(where $\{|a_i\rangle\}$ is some basis on $H_a$). Where did expression (1) come from?, If I knew the expression for $K$, would expression (1) be equivalent to applying $\mathrm{Tr}_b(K)$ = $\sum_{j}\langle b_j|K|b_j\rangle$?

Following the same reasoning, but now in the context of measurements in a composite system ($H = H_{a}\otimes H_{b}$). After measurement of an observable $A = \sum_{a}a|a\rangle\langle a| = \sum_{a}aA_a$ , where $A_a=|a\rangle\langle a|$ is the projector and $\{a\}$ is the discrete spectrum, the state collapses to $$\rho_a = (A_a\otimes I_b)\rho(A_a\otimes I_b)/p_a \tag{2} $$ with $p_a$ the probability of obtaining (a) when a measurement is taken.

Likewise, I would like to know where equation $(2)$ came from. Do equations $(1)$ and $(2)$ have any relationship?

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    $\begingroup$ I'm not sure what exactly you are asking. You want to know where equation (1) came from? And whether $I_a$ is the identity operator on subspace $a$? $\endgroup$
    – noah
    Feb 20, 2021 at 21:07
  • $\begingroup$ yes,i want to know where equation (1) came from? and if identity operator can be written the way I put it? identity operator and the closure relation are the same thing? i would like to know where equation (2) came from? Does Equation 1 and 2 have any relationship? $\endgroup$ Feb 20, 2021 at 21:19
  • $\begingroup$ what definition of partial trace are you going by? $\endgroup$
    – glS
    Feb 22, 2021 at 14:10

1 Answer 1

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A reasonable definition for the partial trace is the following. Given any orthonormal basis sets $|a_i\rangle$ and $|b_i\rangle$ for $\mathcal H_a$ and $\mathcal H_b$ respectively, any operator $K$ on the space $\mathcal H_a\otimes \mathcal H_b$ can be written

$$K = \sum_{ij k\ell} K_{ijk\ell} |a_i\rangle |b_j\rangle \langle a_k|\langle b_\ell|$$

where $K_{ijk\ell} \equiv \langle a_i|\langle b_j| K |a_k\rangle |b_\ell\rangle$, and $\langle a_i|\langle b_j| \equiv \langle a_i|\otimes \langle b_j|$ (I omit the $\otimes$ for notational clarity). The partial trace is then defined to be

$$\mathrm{Tr}_b(K):= \sum_{ik\ell}K_{i\ell k\ell}|a_i\rangle\langle a_k|$$

which is now a linear operator on $\mathcal H_a$ alone, with coefficients $$\bigg(\mathrm{Tr}_b(K)\bigg)_{ik} = \sum_\ell K_{i\ell k\ell}$$

Many people choose to write equation $(1)$ because it gives the impression of tracing over the $|b_i\rangle$ basis while leaving the $|a_i\rangle$ basis alone. If you look too closely at the expression, though, it doesn't really make sense$^\dagger$ - what kind of object is (operator)$\otimes$(vector)?

If I knew the expression for $K$, would expression (1) be equivalent to applying $\mathrm{Tr}_b(K) = \sum_j \langle b_j|K|b_j\rangle$?

That expression doesn't make sense. $K$ acts on $\mathcal H_a\otimes \mathcal H_b$, so what does $K|b_j\rangle$ mean if $|b_j\rangle \in \mathcal H_b$?

Likewise, I would like to know where equation (2) came from.

If you have a pure state $|\psi\rangle\in \mathcal H_a\otimes \mathcal H_b$, then the corresponding density operator is given by $\rho_\psi := \frac{|\psi\rangle \langle \psi|}{\langle \psi|\psi\rangle}$. If you apply a projection operator $P_a\otimes \mathbb I_b$ to your state $|\psi\rangle$, your new density operator becomes

$$\rho = \frac{(P_a \otimes \mathbb I_b)|\psi\rangle\langle \psi|(P_a \otimes \mathbb I_b)}{\langle \psi|(P_a \otimes \mathbb I_b)^2|\psi\rangle} $$

which is equal to your equation (2). In short, density operators inherit their projective evolution from the projective evolution of the states from which they are built. This then extends naturally to states which are not necessarily pure.


$^\dagger$This can be remedied. Objects like these can be defined given a bit of thought, and the result is intuitively just what you'd expect.

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  • $\begingroup$ thank you so much $\endgroup$ Feb 23, 2021 at 22:02

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