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basic_properties_of_linear_operators [2021/02/24 06:55] – [Expectation Values] 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
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   - Let $\ket{e_1}, \ket{e_2}, \cdots$ be an orthonormal basis.  Prove that   - Let $\ket{e_1}, \ket{e_2}, \cdots$ be an orthonormal basis.  Prove that
   \[\sum_j \proj{e_j}\]   \[\sum_j \proj{e_j}\]
-  is the identity operator. +  is the identity operator.\\ \\  
-   +  HINT: You have to prove that $\left ( \sum_j \proj{e_j} \right ) \ket{\psi} = \ket{\psi}$ for all vectors $\ket{\psi}$.  Since $\ket{e_1}, \ket{e_2}, \cdots$ is an orthonormal basis, every vector can be decomposed as $\ket{\psi} = \sum_j b_j \ket{e_j}$ for some coefficients $b_j$.  
-  HINT: You have to prove that $\left ( \sum_j \proj{e_j} \right ) \ket{\psi} = \ket{\psi}$ for all vectors $\ket{\psi}$.  Since $\ket{e_1}, \ket{e_2}, \cdots$ is an orthonormal basis, every vector can be decomposed as $\ket{\psi} = \sum_j b_j \ket{e_j}$ for some coefficients $b_j$. +  - Show that $[\hat{x},\hat{p}] = i\hbar \hat{I}$, where $\hat{p} = -i\hbar \hat{\frac{\D}{\D x}}$.\\ \\ 
-  - Show that $[\hat{x},\hat{p}] = i\hbar \hat{I}$, where $\hat{p} = -i\hbar \hat{\frac{\D}{\D x}}$. +
   HINT: You need to show that $[\hat{x},\hat{p}]\ket{\psi} = i\hbar \ket{\psi}$ for all vectors $\ket{\psi}$.  You may use sloppy notation, i.e.   HINT: You need to show that $[\hat{x},\hat{p}]\ket{\psi} = i\hbar \ket{\psi}$ for all vectors $\ket{\psi}$.  You may use sloppy notation, i.e.
   \[\hat{x} \psi (x) = x\psi(x), \qquad \hat{p} \psi(x) = -i\hbar \frac{\D \psi}{\D x}.\]   \[\hat{x} \psi (x) = x\psi(x), \qquad \hat{p} \psi(x) = -i\hbar \frac{\D \psi}{\D x}.\]