I have a vector space $V$ and a subspace of $V$, $W$. Let $P$ be the projection operator for subspace $W$. Also let the dimension of $W$ be $d$. Also I have two orthonormal basis $(a_1,a_2,...a_d)$ and $(b_1,b_2,....b_d)$ for subspace $W$, where each $a_i$ and $b_i$ $\in V$. Now I can express $P$ in outer product ( bra-ket ) form in the following two ways $$P_1=\sum_{i=1}^{i=d}|a_i\rangle \langle a_i|...(1)$$ $$P_2=\sum_{i=1}^{i=d}|b_i\rangle \langle b_i|...(2)$$ What I know is that both $P_1$ and $P_2$ represent the operator $P$ and are related by $P_1=UP_2U^{\dagger}$ for some unitary operator $U$.
My question is can terms in $P_1$ be re-arranged to get $P_2$. What I mean to say is if I express each $a_i$ in terms of basis $\{b_j\}$ and put it in equation $(1)$ will I get $(2)$?
I am not blindly asking , i tried taking $V$ as 3-D space and $W$ as 2-D space. I was able to show for some examples but in general I am not able to prove.

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    $\begingroup$ Related question by OP: physics.stackexchange.com/q/159131/2451 $\endgroup$
    – Qmechanic
    Commented Jan 14, 2015 at 12:46
  • $\begingroup$ Note that $P_1=P_2$ (not only $P_1=UP_2U^\dagger$). In any case, $U$ is exactly the basis transformation from the $a_i$ to the $b_j$. So what is your question (or, otherwise, where did you get that $U$ you're taking about from)? $\endgroup$ Commented Jan 14, 2015 at 12:52
  • $\begingroup$ @NorbertSchuch yes I understand $P_1$=$P_2$ ,I want to ask whether one representation be obtained from another by re-arranging. The thing you said here that it is same as 0 or (2+3)-5 physics.stackexchange.com/q/159131/2451 $\endgroup$ Commented Jan 14, 2015 at 13:04
  • $\begingroup$ @sasha So is your question "what is $U$", or do you understand what $U$ is? $\endgroup$ Commented Jan 14, 2015 at 13:52
  • $\begingroup$ I understand from Qmechanic's answer what $U$ is, my question is does it mean P1 can be obtained from P2 by rearranging the terms $\endgroup$ Commented Jan 14, 2015 at 13:53

2 Answers 2


Since $\lvert a_i\rangle$ and $\lvert b_i\rangle$ are both bases for the space $W$, there exists a unitary $U=\sum \lvert b_j\rangle \langle a_j\rvert$ which maps $\lvert b_i\rangle=U\lvert a_i\rangle$ for all $i$. This $U$ can be naturally embedded in $V$, i.e., we can think of it as an operator $U:V\rightarrow V$. Then, $$ P_2 = \sum \lvert b_i\rangle\langle b_i\rvert = \sum U \lvert a_i\rangle\langle a_i\rvert U^\dagger = P_1\ . $$

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    $\begingroup$ @sasha Does this answer your question? $\endgroup$ Commented Jan 14, 2015 at 13:55

In terms of the splitting $V=W\oplus W^{\perp}$, the projection operator $P_1=P=P_2$ and unitary operator $U$ are block diagonal

$$ P ~=~ \begin{bmatrix} {\bf 1} & {\bf 0} \\ {\bf 0} & {\bf 0}\end{bmatrix} $$


$$ U ~=~ \begin{bmatrix} U|_W & {\bf 0} \\ {\bf 0} & {\bf 1}\end{bmatrix}, $$

respectively. Clearly, the two operators commute

$$ [P,U]~=~{\bf 0}. $$


$$ UPU^{\dagger}~=~PUU^{\dagger}~=~P. $$

  • $\begingroup$ yes I understood that part from your answer on my previous question.So does it mean $P_1$ can be obtained from $P_2$ by rearranging the terms ? $\endgroup$ Commented Jan 14, 2015 at 13:07
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    $\begingroup$ it depends on what you mean by rearranging the terms. The unitary matrix is encoding an expansion of one base in terms of the other, i.e. $a_i = \sum_k u_{ik}b_k$, so if you replace the $a_i$s with these relations you will get the projection in the base $b_k$. $\endgroup$
    – Phoenix87
    Commented Jan 14, 2015 at 13:59
  • $\begingroup$ yes thats what I wanted to ask thanks for patience , I accept your previous answer , thanks for detailed explanations :) $\endgroup$ Commented Jan 14, 2015 at 14:02

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