2
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It may look a silly question but anybody of you knows what's:

$$(|0\rangle+|1\rangle)\otimes(|0\rangle+|1\rangle)$$ (x: Kronecker operator)

$$(|0\rangle+|1\rangle)*(|0\rangle+|1\rangle)$$ (*: vector multiplication operator)

yielding?

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3
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The Kronecker product of the vectors will lead to

  • $|0\rangle=\begin{pmatrix}1\\0\end{pmatrix}$ and $|1\rangle=\begin{pmatrix}0\\1\end{pmatrix}$

  • $(|0\rangle+|1\rangle)\otimes(|0\rangle+|1\rangle)=|00\rangle+|01\rangle+|10\rangle+|11\rangle=\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}1\\0\end{pmatrix}+\begin{pmatrix}1\\0\end{pmatrix}\otimes\begin{pmatrix}0\\1\end{pmatrix}+\begin{pmatrix}0\\1\end{pmatrix}\otimes\begin{pmatrix}1\\0\end{pmatrix}+\begin{pmatrix}0\\1\end{pmatrix}\otimes\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}+\begin{pmatrix}0\\1\\0\\0\end{pmatrix}+\begin{pmatrix}0\\0\\1\\0\end{pmatrix}+\begin{pmatrix}0\\0\\0\\1\end{pmatrix}=\begin{pmatrix}1\\1\\1\\1\end{pmatrix}$

While the vector multiplication (which I assume that is the Vector inner product) is

  • $(|0\rangle+|1\rangle)*(|0\rangle+|1\rangle)=(|0\rangle+|1\rangle)^\dagger(|0\rangle+|1\rangle)=\langle0|0\rangle+\langle0|1\rangle+\langle1|0\rangle+\langle1|1\rangle=1+0+0+1=2.$
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  • $\begingroup$ It would be better to specify that the bra notation needs to be used for the first term? Else it is not a valid operation. $\endgroup$ – Mahathi Vempati Jun 20 at 8:41
  • 2
    $\begingroup$ It is a valid operation because the dot product specifies such transposition of a vector. That's why I state that I assume the inner product. Obviously if such operation is the standard matrix product, then such operation would not be valid due to dimension mismatch. $\endgroup$ – Josu Etxezarreta Martinez Jun 20 at 9:01
  • $\begingroup$ i see you turned the dirac notation into a normal vector notation, could you specify the resulting vector in dirac notation (or in states notation) ? $\endgroup$ – user1319236 Jun 20 at 9:37
  • $\begingroup$ What do you mean exactly? Those vectors in the bracket notation that are shown in the question are not valid quantum states as they are not normalized, but considering them just as vectors is possible. $\endgroup$ – Josu Etxezarreta Martinez Jun 20 at 10:19
  • $\begingroup$ I meant could you express the results in terms of states ? or in terms of dirac notation ? $\endgroup$ – user1319236 Jun 20 at 10:21

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