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## my Quantum Mechanics cheat sheet

September 8, 2023   Compiled on September 8, 2023 at 5:57pm

 Position Operator $$X$$ Momentum operator $$P$$ Hamilitonian operator $$H$$ Eigenvalue eigenvector relation $$X \ket {x} = x \ket {x}$$ where $$x$$ is the eigenvalue (size) of the $$\ket {x}$$ which is the position vector associated with $$x$$ measured. $$P \ket {\phi _p} = p \ket {\phi _p}$$ where $$p$$ is the momentum of the particle. $$H \ket {\Psi _{E_i}} = E_i \ket {\Psi _{E_i}}$$ where $$E_i$$ is the energy level of the particle. Normalization relation $$\int _{-\infty }^{\infty } \ket {x} \bra {x} \,dx = 1$$ $$\int _{-\infty }^{\infty } \ket {\phi _p} \bra {\phi _p} \,dp = 1$$ $$\int _{-\infty }^{\infty } \ket {\Psi _{E_i}} \bra {\Psi _{E_i}} \,dE = 1$$ orthogonality $$\braket {x|x'}=\delta {(x-x')}$$ $$\braket {\phi _p|\phi _{p'}}=\delta {(p-p')}$$ $$\braket {\Psi _{E_i}|\Psi _{E_j}}=\delta {(E_i-E_j)}$$ Matrix element of operator $$\Braket {x|X|x'}= x' \delta {(x-x')}$$. Operator $$X$$ is diagonal matrix. $$\braket {x|P|x'}=-i \hbar \delta {(x-x')} \frac {d}{d x'}$$ where momentum operator $$P$$ is expressed in position operator $$\ket {x}$$ basis. Note that operator $$P$$ is not a diagonal matrix. $$\Braket {x|H|x'}= ?$$ Function form of the state function $$\ket {\Psi }$$ N/A ? \begin {align*} P \ket {\phi _p} &= p \ket {\phi _p}\\ \int P \ket {x'} \braket {x'|\phi _p} \, dx &= p \int \ket {x'} \braket {x'|\phi _p} \, dx\\ \int \braket {x|P|x'} \braket {x'|\phi _p} \, dx &= p \int \braket {x|x'} \braket {x'|\phi _p} \, dx\\ \int -i \hbar \delta {(x-x')} \frac {d}{d x'} \phi _p(x') \, dx &= p \int \delta {(x-x')} \phi _p(x') \, dx\\ &= \frac {2}{L} \frac {L}{2}\\ &=1 \end {align*} \begin {align*} \braket {\Psi |\Psi } &= \int _{-\infty }^{\infty } \braket {\Psi |x} \braket {x|\Psi } \,dx\\ &= \int _{-\infty }^{\infty } \braket {x|\Psi } \braket {x|\Psi } \,dx\\ &= \int _{-\infty }^{\infty } \Psi ^*(x) \Psi (x) \,dx\\ &= \int _{0}^{L} \left ( \sqrt {\frac {2}{L}} \sin {\frac {n \pi x}{L}} \right )^2 \,dx\\ &= \frac {2}{L} \frac {L}{2}\\ &=1 \end {align*} Vector form to function form $$\Braket {x|\Psi }= \Psi (x)$$ $$\Braket {x|\phi _p}= \phi _p(x)$$ $$\Braket {x|\Psi _{E}}=\Psi _{E}(x)$$ Expansion of state vector $$\ket {\Psi }$$ $$\ket {\Psi } = \int _{-\infty }^{\infty } \ket {x} \braket {x|\Psi } \,dx$$ $$\ket {\Psi } = \int _{-\infty }^{\infty } \ket {\phi _p} \braket {\phi _p|\Psi } \,dp$$ $$\ket {\Psi } = \int _{-\infty }^{\infty } \ket {E_i} \braket {E_i|\Psi } \,di$$ State function $$\ket {\Psi }$$ For inﬁnite potential deep well of width \(x<0