Position Operator $X$

Momentum operator $P$

Hamilitonian operator $H$

$X|x⟩=x|x⟩$ where $x$ is the eigenvalue (size) of the $|x⟩$ which is the position vector associated with $x$ measured.

$P|{\varphi }_p⟩=p|{\varphi }_p⟩$ where $p$ is the momentum of the particle.

$H|{\mathrm{\Psi }}_{E_i}⟩=E_i|{\mathrm{\Psi }}_{E_i}⟩$ where $E_i$ is the energy level of the particle.

Normalization relation

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|x⟩⟨x|dx=1$

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|{\varphi }_p⟩⟨{\varphi }_p|dp=1$

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|{\mathrm{\Psi }}_{E_i}⟩⟨{\mathrm{\Psi }}_{E_i}|dE=1$

orthogonality

$⟨x|x^{\prime }⟩=\delta (x-x^{\prime })$

$⟨{\varphi }_p|{\varphi }_{p^{\prime }}⟩=\delta (p-p^{\prime })$

$⟨{\mathrm{\Psi }}_{E_i}|{\mathrm{\Psi }}_{E_j}⟩=\delta (E_i-E_j)$

$⟨x\left|X\right|x^{\prime }⟩=x^{\prime }\delta (x-x^{\prime })$. Operator $X$ is diagonal matrix.

$⟨x|P|x^{\prime }⟩=-i\hslash \delta (x-x^{\prime })\frac{d}{d{x}^{\prime }}$ where momentum operator $P$ is expressed in position operator $|x⟩$ basis. Note that operator $P$ is not a diagonal matrix.

$⟨x\left|H\right|x^{\prime }⟩=?$

N/A ?

$$\begin{array}{rl}P|{\varphi }_p⟩& =p|{\varphi }_p⟩\\ \int P|x^{\prime }⟩⟨x^{\prime }|{\varphi }_p⟩dx& =p\int |x^{\prime }⟩⟨x^{\prime }|{\varphi }_p⟩dx\\ \int ⟨x|P|x^{\prime }⟩⟨x^{\prime }|{\varphi }_p⟩dx& =p\int ⟨x|x^{\prime }⟩⟨x^{\prime }|{\varphi }_p⟩dx\\ \int -i\hslash \delta (x-x^{\prime })\frac{d}{d{x}^{\prime }}{\varphi }_p(x^{\prime })dx& =p\int \delta (x-x^{\prime }){\varphi }_p(x^{\prime })dx\\ & =\frac{2}{L}\frac{L}{2}\\ & =1\end{array}$$

$$\begin{array}{rl}⟨\mathrm{\Psi }|\mathrm{\Psi }⟩& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}⟨\mathrm{\Psi }|x⟩⟨x|\mathrm{\Psi }⟩dx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}⟨x|\mathrm{\Psi }⟩⟨x|\mathrm{\Psi }⟩dx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{\Psi }}^{\ast }(x)\mathrm{\Psi }(x)dx\\ & ={\int }_0^L{(\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L})}^{2}dx\\ & =\frac{2}{L}\frac{L}{2}\\ & =1\end{array}$$

$⟨x|\mathrm{\Psi }⟩=\mathrm{\Psi }(x)$

$⟨x|{\varphi }_p⟩={\varphi }_p(x)$

$⟨x|{\mathrm{\Psi }}_E⟩={\mathrm{\Psi }}_E(x)$

$|\mathrm{\Psi }⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|x⟩⟨x|\mathrm{\Psi }⟩dx$

$|\mathrm{\Psi }⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|{\varphi }_p⟩⟨{\varphi }_p|\mathrm{\Psi }⟩dp$

$|\mathrm{\Psi }⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|E_i⟩⟨E_i|\mathrm{\Psi }⟩di$

todo

todo

todo

1.

Probability of measuring $x$ given system is in state $|\mathrm{\Psi }⟩$ is $|⟨\mathrm{\Psi }|\mathrm{\Psi }⟩{|}^2$. For infinite potential deep well of width $x<0<L$ this becomes

$$\begin{array}{rl}⟨\mathrm{\Psi }|\mathrm{\Psi }⟩& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}⟨\mathrm{\Psi }|x⟩⟨x|\mathrm{\Psi }⟩dx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}⟨x|\mathrm{\Psi }⟩⟨x|\mathrm{\Psi }⟩dx\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{\Psi }}^{\ast }(x)\mathrm{\Psi }(x)dx\\ & ={\int }_0^L{(\sqrt{\frac{2}{L}}\sin \frac{n\pi x}{L})}^{2}dx\\ & =\frac{2}{L}\frac{L}{2}\\ & =1\end{array}$$

2.

Probability of measuring $x$ given system is in state ${\varphi }_p$

$|\mathrm{\Psi }⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|{\varphi }_p⟩⟨{\varphi }_p|\mathrm{\Psi }⟩dp$

$|\mathrm{\Psi }⟩={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|E_i⟩⟨E_i|\mathrm{\Psi }⟩di$