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In QM I have the state $\lvert 00 \rangle \langle 00 \rvert$. Can anyone tell me what this would look like as a matrix? I know that $$ \lvert 00 \rangle = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}. $$

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$|00\rangle\langle00|$ is a unit matrix element on $\mathbb{C}^2\otimes \mathbb{C}^2$ space. In the same way that $\hat{\boldsymbol{i}}$ or $\hat{\boldsymbol{x}}$ is a unit vector. For example if:

$$M=|00\rangle\langle00|$$

Then this could be represented in matrix form as:

$$M=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0& 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$

In the same way as, if:

$$v = \hat{\boldsymbol{i}}$$

Then,

$$v = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}$$

So the last comment is not true. A ket element represents a vector (not a matrix):

$$|00\rangle\ = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix} \neq \begin{pmatrix} 1&1 \\0&0\end{pmatrix}$$

In general:

  • a ket (e.g. $|00\rangle$) represents a vector
  • a bra (e.g. $\langle00|$) represents a co-vetor, dual-vector or one-form (which all mean the same thing in QM)
  • a ket-bra (e.g. $|00\rangle\langle00|$) a matrix element
  • a bra-ket (e.g. $\langle00||00\rangle = \langle00|00\rangle$) a scalar quantity
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Keep in mind that $|a\rangle=\langle a|^*$. This means that $\langle 00|$ will be the reverse and adjoint of $|00\rangle$. So you'll get $\begin{pmatrix} 1&0 \\1&0\end{pmatrix}$. The product will give you a new $4\times4$ matrix.

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I've never seen this being claimed as true...

$$ \lvert 00 \rangle = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} $$

$\lvert 00 \rangle$ is a shorthand for a kronecker product, $\lvert 0 \rangle \otimes \lvert 0 \rangle$, which should be represented by a vector, not a matrix.

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