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basic_properties_of_linear_operators [2021/02/24 06:57] – [In Class Activities] adminbasic_properties_of_linear_operators [2022/09/27 18:38] (current) – [Examples of Linear Operators] admin
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   The upshot is that the coefficients $\psi(x)$ get transformed to $\psi(-x)$ by the parity operator.  As you may have guessed by now, we will sometimes abuse notation and write   The upshot is that the coefficients $\psi(x)$ get transformed to $\psi(-x)$ by the parity operator.  As you may have guessed by now, we will sometimes abuse notation and write
   \[\hat{\mathcal{P}} \psi(x) = \psi(-x).\]   \[\hat{\mathcal{P}} \psi(x) = \psi(-x).\]
-  * The //**Laplacian operator**// \hat{\nabla}^2:  Although we have not really discussed three-dimensional systems yet, the one-dimensional position basis $\ket{x}$ is generalized to a three-dimensional position basis $\ket{\vec{r}}$ where the basis vectors are now labelled by three-dimensional position vectors $\vec{r}$.  In this basis, a general vector can be decomposed as+  * The //**Laplacian operator**// $\hat{\nabla}^{2}$:  Although we have not really discussed three-dimensional systems yet, the one-dimensional position basis $\ket{x}$ is generalized to a three-dimensional position basis $\ket{\vec{r}}$ where the basis vectors are now labelled by three-dimensional position vectors $\vec{r}$.  In this basis, a general vector can be decomposed as
   \[\ket{\psi} = \int \psi(\vec{r}) \ket{\vec{r}}\,\D V,\]   \[\ket{\psi} = \int \psi(\vec{r}) \ket{\vec{r}}\,\D V,\]
   where the components $\psi(\vec{r})$ are now a scalar function of the position vector and $\D V$ is the three-dimensional volume element $\D V = \D x\D y\D z$.  The Laplacian operator then acts as   where the components $\psi(\vec{r})$ are now a scalar function of the position vector and $\D V$ is the three-dimensional volume element $\D V = \D x\D y\D z$.  The Laplacian operator then acts as