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basic_properties_of_linear_operators [2021/02/24 06:48] – [The Dirac Notaty] 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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 \sand{\psi}{\hat{x}}{\psi} & = \int_{-\infty}^{+\infty} \D x' \, \psi^*(x') \bra{x'} \int_{-\infty}^{+\infty} \D x \,\ x \psi(x) \ket{x} \\ \sand{\psi}{\hat{x}}{\psi} & = \int_{-\infty}^{+\infty} \D x' \, \psi^*(x') \bra{x'} \int_{-\infty}^{+\infty} \D x \,\ x \psi(x) \ket{x} \\
 & = \int_{-\infty}^{+\infty} \D x' \, \int_{-\infty}^{+\infty} \D x \, x \psi^*(x')\psi(x) \braket{x'}{x} \\ & = \int_{-\infty}^{+\infty} \D x' \, \int_{-\infty}^{+\infty} \D x \, x \psi^*(x')\psi(x) \braket{x'}{x} \\
-& = \int_{-\infty}^{+\infty} \D x' \, \int_{-\infty}^{+\infty} \D x \, x \psi^*(x')\psi(x) \delta(x-x') \\+& = \int_{-\infty}^{+\infty} \D x' \, \int_{-\infty}^{+\infty} \D x \, x \psi^*(x')\psi(x) \delta(ex-x') \\
 & = \int_{-\infty}^{+\infty} \D x \, x\psi^*(x)\psi(x) \\ & = \int_{-\infty}^{+\infty} \D x \, x\psi^*(x)\psi(x) \\
 & = \int_{-\infty}^{+\infty} x\Abs{\psi(x)}^2 \, \D x. & = \int_{-\infty}^{+\infty} x\Abs{\psi(x)}^2 \, \D x.
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 We will see later in the course that other physical quantities are also represented by linear operators.  If $\hat{A}$ represents such a physical quantity then $\sand{\psi}{\hat{A}}{\psi}$ is its expectation value in the sense of probability theory. We will see later in the course that other physical quantities are also represented by linear operators.  If $\hat{A}$ represents such a physical quantity then $\sand{\psi}{\hat{A}}{\psi}$ is its expectation value in the sense of probability theory.
 +
 +{{:question-mark.png?direct&50|}}
 +====== In Class Activities ======
 +
 +  - Let $\ket{e_1}, \ket{e_2}, \cdots$ be an orthonormal basis.  Prove that
 +  \[\sum_j \proj{e_j}\]
 +  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 $b_j$. 
 +  - 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.
 +  \[\hat{x} \psi (x) = x\psi(x), \qquad \hat{p} \psi(x) = -i\hbar \frac{\D \psi}{\D x}.\]