I How Do You Compute the Density Matrix of a Bipartite State?

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To compute the density matrix of a bipartite state, the simplest method is to use the outer product of the state vector with itself, expressed as ρ = |φ><φ|. Converting the state to matrix form can complicate the calculation, particularly when determining the dual vector (bra) of the state. The confusion arises from how the complex conjugate transpose transforms a column vector into a matrix. The consensus is that directly computing the density matrix from the state vector is the most straightforward approach. Thus, ρ = |φ><φ| is the recommended method.
Rayan
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What is the easiest way to compute a density matrix of bipartite states?
If we for example have such a bipartite state:

$$ | \phi > = \frac{1}{2} [ |0>|0> + |1>|0> + |0>|1> + |1>|1> ] $$

What is the easiest way to compute a density matrix of bipartite states? Should I just compute it as it is? i.e:

$$ \rho = | \phi > < \phi | $$

Or should I convert to matrix form first? Any advice appreciated!

I tried to convert it to matrix form and got the following:

$$ | \phi > = \frac{1}{2}
\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix} $$

and

$$ < \phi | = \frac{1}{2}
\begin{pmatrix}
1 & 1\\
1 & 1
\end{pmatrix}
$$

But then I don't think it is possible to compute the following outer product?

$$ \rho = \frac{1}{4}
\begin{bmatrix}
1 \\
1 \\
1 \\
1
\end{bmatrix} \cdot
\begin{pmatrix}
1 & 1\\
1 & 1
\end{pmatrix}
$$
 
Last edited:
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Things went wonky when you calculated the dual vector (bra) of the state. How did the complex conjugate transpose turn a column vector into a matrix?
 
Rayan said:
Should I just compute it as it is? i.e:
$$ \rho = | \phi > < \phi | $$
Yes.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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