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adjoints_and_hermitian_operators [2022/09/27 20:50] – [Projection Operators] adminadjoints_and_hermitian_operators [2022/10/06 00:45] (current) – [Projection Operators] admin
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 where $V^{\perp}$ is the orthogonal complement of $V$ in $\mathcal{H}$. where $V^{\perp}$ is the orthogonal complement of $V$ in $\mathcal{H}$.
  
-In order to completely define the operator, we need to say what it does to vectors that are neither in $V$ or $V^{\perp}$.  However, since $\mathcal{H} = V\oplus V^{\perp}$, we can write any vector $\ket{\psi}$ as $\ket{\psi} = a\ket{\phi} + b\ket{\phi^{\perp}}$, where $\ket{\psi}\in V$ and $\ket{\psi^{\perp}}\in V^{\perp}$.  Then, if we want $\hat{P}_V$ to be a //linear// operator, we must have+In order to completely define the operator, we need to say what it does to vectors that are neither in $V$ or $V^{\perp}$.  However, since $\mathcal{H} = V\oplus V^{\perp}$, we can write any vector $\ket{\psi}$ as $\ket{\psi} = a\ket{\phi} + b\ket{\phi^{\perp}}$, where $\ket{\phi}\in V$ and $\ket{\phi^{\perp}}\in V^{\perp}$.  Then, if we want $\hat{P}_V$ to be a //linear// operator, we must have
 \[\hat{P}_V \ket{\psi} = \hat{P}_V \left ( a\ket{\phi} + b\ket{\phi^{\perp}} \right ) = a\hat{P}_V\ket{\phi} + b\hat{P}_V \ket{\phi^{\perp}} = a\ket{\phi}.\] \[\hat{P}_V \ket{\psi} = \hat{P}_V \left ( a\ket{\phi} + b\ket{\phi^{\perp}} \right ) = a\hat{P}_V\ket{\phi} + b\hat{P}_V \ket{\phi^{\perp}} = a\ket{\phi}.\]
 In other words, the projection operator $\hat{P}_V$ simply returns the component of the vector $\ket{\psi}$ that lies in the subspace $V$. In other words, the projection operator $\hat{P}_V$ simply returns the component of the vector $\ket{\psi}$ that lies in the subspace $V$.
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 Now, the terms $\sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}}, \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2}$ and $\sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}}$ are all zero because $\ket{\phi_1^{\perp}}$ and $\ket{\phi_2^{\perp}}$ are in a subspace orthogonal to $V$, so $\hat{P}_V\ket{\phi_1^{\perp}} = \hat{P}_V \ket{\phi_2^{\perp}} = 0$.  In particular, since the complex conjugate of zero is zero, this means that Now, the terms $\sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}}, \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2}$ and $\sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}}$ are all zero because $\ket{\phi_1^{\perp}}$ and $\ket{\phi_2^{\perp}}$ are in a subspace orthogonal to $V$, so $\hat{P}_V\ket{\phi_1^{\perp}} = \hat{P}_V \ket{\phi_2^{\perp}} = 0$.  In particular, since the complex conjugate of zero is zero, this means that
 \begin{align*} \begin{align*}
-  \sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}} & = \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1}^*, & \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2} & = \sand{\phi_2}{\hat{P}_V}{\phi_1^{\perp}}^*, \\ +  \sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}} & = \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1}^*, & 
-   \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}} & = \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1^{\perp}.+  \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2} & = \sand{\phi_2}{\hat{P}_V}{\phi_1^{\perp}}^*, & 
 +   \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}} & = \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1^{\perp}}^*.
 \end{align*} \end{align*}
 In addition In addition
 \begin{align*} \begin{align*}
    \sand{\phi_1}{\hat{P}_V}{\phi_2} & = \braket{\phi_1}{\phi_2} \\    \sand{\phi_1}{\hat{P}_V}{\phi_2} & = \braket{\phi_1}{\phi_2} \\
-   & = \braket{\phi_2}{\phi_1}^* +   & = \braket{\phi_2}{\phi_1}^* \\ 
-   & = \sand{\phi_2}{\hat{P}_V}{\sand{\phi_1}}^*.+   & = \sand{\phi_2}{\hat{P}_V}{\phi_1}^*.
 \end{align*} \end{align*}
 This is because $\ket{\phi_1},\ket{\phi_2}\in V$, so $\hat{P}_V \ket{\phi_1} = \ket{\phi_1}$ and $\hat{P}_V \ket{\phi_2} = \ket{\phi_2}$. This is because $\ket{\phi_1},\ket{\phi_2}\in V$, so $\hat{P}_V \ket{\phi_1} = \ket{\phi_1}$ and $\hat{P}_V \ket{\phi_2} = \ket{\phi_2}$.
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 \begin{align*} \begin{align*}
   \sand{\psi_1}{\hat{P}_V}{\psi_2} & =\sand{\phi_1}{\hat{P}_V}{\phi_2} + \sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}} + \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2} + \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}} \\   \sand{\psi_1}{\hat{P}_V}{\psi_2} & =\sand{\phi_1}{\hat{P}_V}{\phi_2} + \sand{\phi_1}{\hat{P}_V}{\phi_2^{\perp}} + \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2} + \sand{\phi_1^{\perp}}{\hat{P}_V}{\phi_2^{\perp}} \\
-  & = \sand{\phi_2}{\hat{P}_V}{\phi_1}^* + \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1}^* + \sand{\phi_2}{\hat{P}_V}{\phi_1^{\perp}}^* + \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1^{\perp}}^* +  & = \sand{\phi_2}{\hat{P}_V}{\phi_1}^* + \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1}^* + \sand{\phi_2}{\hat{P}_V}{\phi_1^{\perp}}^* + \sand{\phi_2^{\perp}}{\hat{P}_V}{\phi_1^{\perp}}^* \\ 
-  & = \left [ \left (  \bra{\phi_2} + \bra{\phi_2^{\perp}}\right ) \hat{P}_V \left ( \ket{\phi_1} + \ket{\phi_1^{\perp}}\right ) right ]^*+  & = \left [ \left (  \bra{\phi_2} + \bra{\phi_2^{\perp}}\right ) \hat{P}_V \left ( \ket{\phi_1} + \ket{\phi_1^{\perp}}\right ) \right ]^* \\
   & = \sand{\psi_2}{\hat{P}_V}{\psi_1}^*,   & = \sand{\psi_2}{\hat{P}_V}{\psi_1}^*,
 \end{align*} \end{align*}
 so $\hat{P}_V^{\dagger} = \hat{P}_V$. so $\hat{P}_V^{\dagger} = \hat{P}_V$.
 +
 +To prove that projection operators are idempotent, write an arbitrary vector $\ket{\psi}$ as
 +\[\ket{\psi} = \ket{\phi} + \ket{\phi^{\perp}},\]
 +with $\ket{\phi} \in V$ and $\ket{\phi^{\perp}} \in V^{\perp}$.  Then we have
 +\[\hat{P}_V \ket{\psi} = \ket{\phi},\]
 +but also
 +\[\hat{P}_V^2 \ket{\psi} = \hat{P}_V \hat{P}_V \ket{\psi} = \hat{P_V} \ket{\phi} = \ket{\phi},\]
 +since $\ket{\phi} \in V$.  Since this is true for an arbitrary vector $\ket{\psi}$, we have $\hat{P}_V^2 = \hat{P}_V$.
  
 We conclude with some more basic properties of projectors. We conclude with some more basic properties of projectors.