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

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SUMMARY

The discussion focuses on computing the density matrix of a bipartite quantum state represented as $$ | \phi > = \frac{1}{2} [ |0>|0> + |1>|0> + |0>|1> + |1>|1> ] $$ The consensus is that the density matrix can be computed directly using the outer product $$ \rho = | \phi > < \phi | $$ without the need to convert the state to matrix form first. The participants confirm that the complexity arises from calculating the dual vector, but the direct computation method is sufficient and recommended.

PREREQUISITES
  • Understanding of quantum states and notation, specifically bra-ket notation.
  • Familiarity with density matrices in quantum mechanics.
  • Knowledge of outer products and their application in quantum theory.
  • Basic linear algebra concepts, including vector and matrix operations.
NEXT STEPS
  • Study the properties of density matrices in quantum mechanics.
  • Learn about the outer product and its significance in quantum state representation.
  • Explore the concept of dual vectors and their role in quantum mechanics.
  • Investigate examples of bipartite quantum states and their density matrices.
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Quantum physicists, students of quantum mechanics, and anyone interested in understanding the computation of density matrices for bipartite states.

Rayan
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TL;DR
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?
 
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Likes   Reactions: PeterDonis
Rayan said:
Should I just compute it as it is? i.e:
$$ \rho = | \phi > < \phi | $$
Yes.
 

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