Showing that a state is unentangled under a certain condition

[JD_PM]

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

 

[PeroK]

To show implication one way, you could start with an unentangled state:
$$(a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle)$$
And show that the equation holds for the resulting coefficients.

The reverse implication requires a bit of fiddling about with complex numbers to obtain $a, b, c, d$.

 

[mfb]

The reverse implication requires a bit of fiddling about with complex numbers to obtain $a, b, c, d$.

Which is still far easier than the giant equations OP set up.

 

[JD_PM]

Before proceeding: my understanding of an unentangled state is the following

$$|\phi \rangle = \sum_{m,n} c_{m,n} | a_m b_n \rangle=\sum_{m,n} w_m l_n | a_m \rangle \otimes| b_n \rangle$$

i.e. each state ($| a_m \rangle, | b_n \rangle$) has its own complex coefficient ($w_m, l_n$).

To show implication one way, you could start with an unentangled state:
$$(a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle)$$
And show that the equation holds for the resulting coefficients.

Alright so we have

$$(a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle) = ac|00\rangle + ad|01\rangle + bc|10\rangle + bd |11\rangle$$

But here's my confusion: why does $acbd = adbc$ condition imply unentanglement?

 

[PeroK]

Before proceeding: my understanding of an unentangled state is the following

$$|\phi \rangle = \sum_{m,n} c_{m,n} | a_m b_n \rangle=\sum_{m,n} w_m l_n | a_m \rangle \otimes| b_n \rangle$$

i.e. each state ($| a_m \rangle, | b_n \rangle$) has its own complex coefficient ($w_m, l_n$).
Alright so we have

$$(a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle) = ac|00\rangle + ad|01\rangle + bc|10\rangle + bd |11\rangle$$

But here's my confusion: why does $acbd = adbc$ condition imply unentanglement?

A state is not entangled iff it is the product of (pure) single-particle states. That means explicitly that:
$$|\phi \rangle = (a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle)$$
Where $$a|0\rangle + b|1\rangle, \ \ \text{and} \ \ c|0\rangle + d|1\rangle$$ are the single-particle states. The state is entangled, therefore, iff it cannot be written in this form.

You have shown the implication one way: unentangled implies $c_{00}c_{11} = c_{01}c_{10}$.

The converse requires some work. You could start with this equation and explicitly construct $a, b, c, d$ from the coefficients $c_{mn}$.

 

[JD_PM]

The converse requires some work. You could start with this equation and explicitly construct $a, b, c, d$ from the coefficients $c_{mn}$.

Let's proceed.

$\Rightarrow$ We assume $c_{00}c_{11} = c_{01}c_{10}$.

Based on $|\phi \rangle = \sum_{m,n} c_{m,n} | a_m b_n \rangle$ we get

$$|\phi \rangle = c_{00} | 00 \rangle + c_{11} | 11 \rangle + c_{10} | 10 \rangle + c_{01} | 01 \rangle $$

We now can set $c_{00}=ac, c_{11}=bd, c_{10}=bc, c_{01}=ad$ to get

$$|\phi \rangle = (a|0\rangle + b|1\rangle) (c|0\rangle + d|1\rangle)$$

I guess we now have to use $c_{00}c_{11} = c_{01}c_{10}$ to show that $(a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle)$ holds.

Mmm I still do not see it though ...

 

[PeroK]

Let's proceed.

$\Rightarrow$ We assume $c_{00}c_{11} = c_{01}c_{10}$.

Based on $|\phi \rangle = \sum_{m,n} c_{m,n} | a_m b_n \rangle$ we get

$$|\phi \rangle = c_{00} | 00 \rangle + c_{11} | 11 \rangle + c_{10} | 10 \rangle + c_{01} | 01 \rangle $$

We now can set $c_{00}=ac, c_{11}=bd, c_{10}=bc, c_{01}=ad$ to get

$$|\phi \rangle = (a|0\rangle + b|1\rangle) (c|0\rangle + d|1\rangle)$$

I guess we now have to use $c_{00}c_{11} = c_{01}c_{10}$ to show that $(a|0\rangle + b|1\rangle) \otimes (c|0\rangle + d|1\rangle)$ holds.

Mmm I still do not see it though ...

You need to use the constraints of the problem.

First, if any of the coefficents $c_{mn} = 0$, then the problem becomes trivial (exercise). So, the difficult case is when they are all non-zero. In that case, you have $c_{11} = \frac{c_{01}c_{10}}{c_{00}}$.

You may also need that $c_{00}^2 + c_{01}^2 + c_{10}^2 + c_{11}^2 = 1$.

Finally, it helps a little if you see that you can always take $a$ to be real.

Then, it's not too bad. But, it's not something you can just "see".

 

[JD_PM]

Mmm honestly I still do not see it.

I am sure it has to be done by comparing $c_{mn}$ to $a,b,c,d$ coefficients.

If I get it I'll post it

 

[PeroK]

Mmm honestly I still do not see it.

I am sure it has to be done by comparing $c_{mn}$ to $a,b,c,d$ coefficients.

If I get it I'll post it

Try
$$a = \frac{|c_{00}|}{\sqrt{|c_{00}|^2 + |c_{10}|^2}}$$

 

[JD_PM]

Hi PeroK

Try
$$a = \frac{|c_{00}|}{\sqrt{|c_{00}|^2 + |c_{10}|^2}}$$

Naive question: how did you get it?

 

[PeroK]

Naive question: how did you get it?

First, let me use $A, B, C, D$ instead of $c_{00} \dots$. What I did was this:
$$ ac = A, ad = B, bc = C, bd = D = \frac{BC}{A}$$
This gives
$$\frac b a = \frac C A, \ \ \text{hence} \ \ b = \frac C A a$$
Then use
$$|a|^2 + |b|^2 = 1$$
To get $|a|$ and then take $a$ to be real and positive.