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        <title>de_broglie_matter_waves - [1.v.1 de Broglie Wavelength and Wave Vector] </title>
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        <description>1.v.1 de Broglie Wavelength and Wave Vector

The photoelectric effect and Compton scattering show that electromagnetic waves sometimes exhibit particle-like properties.  In 1923, Louis de Broglie proposed that matter, which we normally think of as made up of particles like electrons, protons and neutrons, should have wave-like properties.  In other words, \(E = pc\)$E = h\nu$$p = h\nu /c = h/\lambda$$\lambda = h/p$$p$\[\boxed{\lambda = \frac{h}{p}.}\]$\vec{p}$$\vec{k}$$k = \frac{2\pi}{\lambda}$\…</description>
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        <description>In 1923, Arthur Compton found that the wavelength of X-rays scattered off free electrons was larger than the wavelength of the incident radiation.  Classically, X-rays are too energetic to be absorbed by free electrons.  They would provide an oscillating electromagnetic field that would cause the electrons to oscillate at the same frequency.  Therefore, they should emit X-rays with the same frequency as the incident radiation.$E=h\nu$\[\vec{p} = \vec{p} + \vec{P}_e \qquad\qquad \Rightarrow \qqua…</description>
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        <description>Proof of the Spectral Theorem for Normal Operators in Finite Dimensions

As a reminder, a normal operator is an operator $\hat{A}$ that satisfies $[\hat{A},\hat{A}^{\dagger}] = 0$, and the spectral theorem is as follows.

Theorem
Let $a_1,a_2,\cdots$ be the (distinct) eigenvalues of an operator $\hat{A}$.  Then, $\hat{A}$ is a normal operator if an only if all of the following hold:$\hat{P}_{a_j}\hat{P}_{a_k} = \delta_{jk}\hat{P}_{a_j}$$\sum_j \hat{P}_{a_j} = \hat{I}$$\hat{A} = \sum_j a_j \hat{P…</description>
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        <description>A vector $\ket{\psi}$ is called an eigenvector (also called an eigenstate in quantum mechanics) of an operator $\hat{A}$ if
\[\hat{A} \ket{\psi} = a \ket{\psi},\]
where $a$ is a scalar called an eigenvalue of $\hat{A}$.

As an example, all vectors are eigenvectors of the identity operator $\hat{I}$ with eigenvalue $1$, since
\[\hat{I} \ket{\psi} = \ket{\psi} = 1\ket{\psi}.\]

The importance of eigenvalues and eigenvectors is that $\ket{a}$$\hat{A}$$a$$\sand{a}{\hat{A}}{a}$$\hat{A}$\[\sand{a}{\ha…</description>
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        <title>functions_inverses_and_unitary_operators - [Interaction of Functions with Commutators] </title>
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        <description>Functions of Operators

Let $f$ be a (complex) function with Taylor expansion
\[f(z) = \sum_{n=0}^{\infty} a_n z^n,\]
and radius of convergence $|z| \leq r$.

We can extend $f$ to be a function on linear operators by defining 
\[f(\hat{A}) = \sum_{n=0}^{\infty} a_n \hat{A}^n.\]

It is possible to prove that this series converges if
\[\sup_{\{\ket{\psi}| \| \psi \| = 1 \}} \Abs{\sand{\psi}{\hat{A}}{\psi}} \leq r.\]

Interaction of Functions with Commutators
$[\hat{A}+\hat{B},\hat{C}] = [\hat{A},\…</description>
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        <description>Commutators

The commutator of two operators $\hat{A}$ and $\hat{B}$ is
\[\boxed{[\hat{A},\hat{B} ] = \hat{A}\hat{B}-\hat{B}\hat{A}.}\]

Two operators commute if
\[[\hat{A},\hat{B}] = 0,\]
or, equivalently
\[\hat{A}\hat{B} = \hat{B}\hat{A}.\]

Clearly, an operator always commutes with itself $[\hat{A},\hat{A}] = 0$.

A little less obviously, if $\hat{A}$, $\hat{B}$ and $\hat{A}\hat{B}$ are all Hermitian then $\hat{A}$ and $\hat{B}$ commute.  Here is the proof:
\begin{align*}
[\hat{A},\hat{B}] &amp; …</description>
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