I would like to solve the following Task. Consider the states

$|\psi_1\rangle=\cos(\Theta_1/2)|0\rangle+\sin(\Theta_1/2)e^{i \phi_1}|1\rangle$

$|\psi_2\rangle=\cos(\Theta_2/2)|0\rangle+\sin(\Theta_2/2)e^{i \phi_2}|1\rangle$

$|\phi\rangle=\frac{1}{\sqrt{3}}(|0\rangle+ |1\rangle+ |2\rangle)$

and the operator $P_0$ acting in the hilbert space $\mathbb{C^3}\bigotimes\mathbb{C^4}$

$P_0=|0\rangle\langle 0| \bigotimes(|1\rangle \bigotimes |0\rangle\langle1| \bigotimes \langle0|)$.

Now, calculate the probability

$p_0=(\langle \phi|\bigotimes \langle \psi |)P_0(|\phi\rangle \bigotimes |\psi\rangle)$ with $|\psi\rangle=|\psi_1\rangle \bigotimes |\psi_2\rangle$.

The Point is, I don't know how to apply $P_0$ on $(|\phi\rangle \bigotimes |\psi\rangle$.

How does $(|1\rangle \bigotimes |0\rangle\langle1| \bigotimes \langle0|)$ act on $|\psi_1\rangle \bigotimes |\psi_2\rangle$? Those three tensor products are confusing me. So far, I have only seen operators where the number of tensor products in the operator is the same as the number of tensor products in the state it is acting on.

I know how to calculate it in matrix representation, but I don't know how to solve this with dirac-notation. How does this work?

Btw how do you find out in this case what the dimensions of the mapping $P_0$ are? I know it is mapping from $\mathbb{C^4}\rightarrow \mathbb{C^4}$, but how do find this out in a mathematically correct way?


  • $\begingroup$ Okay, forget the operator for a second. Do you know how to act with $(\langle 1 | \otimes \langle 0 |)$ on $|\psi_1 \rangle \otimes |\psi_2 \rangle$? That's all you need to know. $\endgroup$ – knzhou Sep 11 '16 at 5:39
  • $\begingroup$ yes, i know how this is done. But I don't see how this could help me $\endgroup$ – anonymous Sep 11 '16 at 5:41
  • $\begingroup$ This quantity is a number. Then $|1 \rangle \otimes |0 \rangle$ acts on this number. That's it! $\endgroup$ – knzhou Sep 11 '16 at 5:42
  • $\begingroup$ but doesn't it make a difference if it's $(|0\rangle \otimes |1\rangle)(\langle 0| \otimes \langle1 |)$ or $(|0\rangle \otimes |1\rangle\langle 0| \otimes \langle1 |)$? (the difference are the parentheses) $\endgroup$ – anonymous Sep 11 '16 at 5:48
  • $\begingroup$ okay, but in my task it says $(|0\rangle \otimes |1\rangle\langle 0| \otimes \langle1 |)$. btw why is this 1? $\endgroup$ – anonymous Sep 11 '16 at 5:51

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