- #1
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- Homework Statement
- Show that a two-qubit state vector
$$ |\phi \rangle = c_{00} |00\rangle + c_{01} |01\rangle + c_{10} |10\rangle + c_{11} |11\rangle \tag{*}$$
is non-entangled iff
$$c_{00} c_{11} = c_{01}c_{10} \tag{**}$$
- Relevant Equations
- $$\hat \rho = \sum_m \rho_m |\phi_m \rangle \langle \phi_m | \tag{1}$$
$$|\phi_k \rangle = \sum_{m,n} c_{m,n}^k | a_m b_n \rangle \tag{2}$$
More below
This is an iff statement, so we proceed as follows
##\Rightarrow## We assume that ##|\phi \rangle## is uncorrelated. Thus the state operator must be of the form ##\hat \rho = \rho^{(1)} \otimes \rho^{(2)}## (equation ##8.16## in Ballentine's book).
The spectral decomposition of the state operator ##\hat \rho## is
$$\hat \rho = \sum_m \rho_m |\phi_m \rangle \langle \phi_m | \tag{1}$$
The eigenvectors of ##\hat \rho## can be expanded in terms of its basis vectors (as suggested by ##(*)##)
$$|\phi_k \rangle = \sum_{m,n} c_{m,n}^k | a_m b_n \rangle \tag{2}$$
Plugging ##(2)## into ##(1)## we get
$$\rho = \sum_k \rho_k \sum_{m,n} \sum_{m',n'} \Big( c_{m,n}^k \Big) \Big( c_{m',n'}^k \Big)^* | a_m b_n \rangle \langle a_{m'} b_{n'} | \tag{3}$$
By definition we know that
$$\rho^{(1)} := \text{Tr}^{(2)} \rho \tag{4}$$
The matrix elements of ##\rho^{(1)}## are (i.e. we sum over ##n##)
$$\langle a_m | \rho^{(1)} | a_{m'} \rangle = \sum_n \langle a_m b_n| \rho | a_{m'} b_{n} \rangle \tag{5}$$
Based on ##(3),(5)## we get that the partial state ##\rho^{(1)}## has the following form
$$\rho^{(1)} = \sum_k \rho_k \sum_{m,m'} \sum_{n} \Big( c_{m,n}^k \Big) \Big( c_{m',n}^k \Big)^* | a_m \rangle \langle a_{m'}| \tag{6}$$
Analogously we get that ##\rho^{(2)}## has the following form
$$\rho^{(2)} = \sum_k \rho_k \sum_{n ,n'} \sum_{m} \Big( c_{m,n}^k \Big) \Big( c_{m,n'}^k \Big)^* | b_n \rangle \langle b_{n'}| \tag{7}$$
It looks to me that the statement is assuming that ##\rho_k=1## for all values of ##k## (where ##k=1,2,3,4##).
As the state operator must be of the form ##\hat \rho = \rho^{(1)} \otimes \rho^{(2)}## we get
$$\hat \rho = \Big[\sum_{m,m'} \sum_{n} \Big( c_{m,n} \Big) \Big( c_{m',n} \Big)^* | a_m \rangle \langle a_{m'}|\Big] \otimes \Big[\sum_{n ,n'} \sum_{m} \Big( c_{m,n} \Big) \Big( c_{m,n'} \Big)^* | b_n \rangle \langle b_{n'}|\Big]$$ $$=\Big[\sum_{m,m'} \sum_{n} \Big( c_{m,n} \Big) \Big( c_{m',n} \Big)^* \Big]\Big[\sum_{n ,n'} \sum_{m} \Big( c_{m,n} \Big) \Big( c_{m,n'} \Big)^* \Big]|a_{m} b_{n} \rangle \langle a_{m'} b_{n'}| \tag{8}$$
The issue I have is that I still do not see how to show ##c_{00} c_{11} = c_{01}c_{10}## out of ##(8)##
Any help is appreciated.
Thank you
##\Rightarrow## We assume that ##|\phi \rangle## is uncorrelated. Thus the state operator must be of the form ##\hat \rho = \rho^{(1)} \otimes \rho^{(2)}## (equation ##8.16## in Ballentine's book).
The spectral decomposition of the state operator ##\hat \rho## is
$$\hat \rho = \sum_m \rho_m |\phi_m \rangle \langle \phi_m | \tag{1}$$
The eigenvectors of ##\hat \rho## can be expanded in terms of its basis vectors (as suggested by ##(*)##)
$$|\phi_k \rangle = \sum_{m,n} c_{m,n}^k | a_m b_n \rangle \tag{2}$$
Plugging ##(2)## into ##(1)## we get
$$\rho = \sum_k \rho_k \sum_{m,n} \sum_{m',n'} \Big( c_{m,n}^k \Big) \Big( c_{m',n'}^k \Big)^* | a_m b_n \rangle \langle a_{m'} b_{n'} | \tag{3}$$
By definition we know that
$$\rho^{(1)} := \text{Tr}^{(2)} \rho \tag{4}$$
The matrix elements of ##\rho^{(1)}## are (i.e. we sum over ##n##)
$$\langle a_m | \rho^{(1)} | a_{m'} \rangle = \sum_n \langle a_m b_n| \rho | a_{m'} b_{n} \rangle \tag{5}$$
Based on ##(3),(5)## we get that the partial state ##\rho^{(1)}## has the following form
$$\rho^{(1)} = \sum_k \rho_k \sum_{m,m'} \sum_{n} \Big( c_{m,n}^k \Big) \Big( c_{m',n}^k \Big)^* | a_m \rangle \langle a_{m'}| \tag{6}$$
Analogously we get that ##\rho^{(2)}## has the following form
$$\rho^{(2)} = \sum_k \rho_k \sum_{n ,n'} \sum_{m} \Big( c_{m,n}^k \Big) \Big( c_{m,n'}^k \Big)^* | b_n \rangle \langle b_{n'}| \tag{7}$$
It looks to me that the statement is assuming that ##\rho_k=1## for all values of ##k## (where ##k=1,2,3,4##).
As the state operator must be of the form ##\hat \rho = \rho^{(1)} \otimes \rho^{(2)}## we get
$$\hat \rho = \Big[\sum_{m,m'} \sum_{n} \Big( c_{m,n} \Big) \Big( c_{m',n} \Big)^* | a_m \rangle \langle a_{m'}|\Big] \otimes \Big[\sum_{n ,n'} \sum_{m} \Big( c_{m,n} \Big) \Big( c_{m,n'} \Big)^* | b_n \rangle \langle b_{n'}|\Big]$$ $$=\Big[\sum_{m,m'} \sum_{n} \Big( c_{m,n} \Big) \Big( c_{m',n} \Big)^* \Big]\Big[\sum_{n ,n'} \sum_{m} \Big( c_{m,n} \Big) \Big( c_{m,n'} \Big)^* \Big]|a_{m} b_{n} \rangle \langle a_{m'} b_{n'}| \tag{8}$$
The issue I have is that I still do not see how to show ##c_{00} c_{11} = c_{01}c_{10}## out of ##(8)##
Any help is appreciated.
Thank you