What are the physical meanings of the outer product when writing expressions for unitary gates?


I'm really confused with the interpretation of those equations:

$1.$ The evolution of states under unitary operations can be expressed as $$ U = \sum_k\exp(i\phi_k)|\psi_k\rangle\langle\psi_k| $$

$2.$ The controlled operation in quantum computing is defined as $$ |0\rangle\langle0|\otimes I + |1\rangle\langle1|\otimes e^{i\alpha}I = \begin{bmatrix} 1 & 0\\ 0 & e^{i\alpha} \end{bmatrix} \otimes I $$

I'm not pretty sure how to understand them. Also, the outer product is present in both equations, so I'm wondering could those equations be explained in terms of the projection operator? What are the physical meanings of the outer product?



Posted 2020-11-01T04:29:58.150

Reputation: 988

Please link to the source where you found this. – keisuke.akira – 2020-11-01T05:30:19.797

@keisuke.akira Thanks for the answer, the link I found this is https://www.youtube.com/watch?v=iLcQ-X6QzvU

– Zhengrong – 2020-11-01T14:31:57.690



  1. This is simply the spectral decomposition of a unitary operator. The eigenvalues of a unitary operator are phases, i.e., of the form $e^{i \theta}$. And the rank-1 projectors are the eigenprojectors (again from the spectral theorem) and have the form $| \psi_{k} \rangle \langle \psi_{k} |$ for a non-degenerate spectrum (otherwise, their rank is greater than one and they cannot be written as the outer product of a state vector).

  2. This is clearly wrong. The action of this is to provide a trivial global phase. I think you mean something like $|0\rangle\langle 0|\otimes I+| 1\rangle\langle 1| \otimes R_z(\alpha)$, where $R_z(\alpha)$ is a phase-shift gate? Or more generally, any two-qubit controlled-$U$ gate can be written as $|0\rangle\langle 0|\otimes I+| 1\rangle\langle 1| \otimes U$.

Does this help?


Posted 2020-11-01T04:29:58.150

Reputation: 1 383

2The action of 2 is not trivial. It is a phase-shift gate on the ‘left’ qubit. – tsgeorgios – 2020-11-01T07:38:24.660