Und das ist die Sandbox
Mathematik-Hilfe [[1]]
3 2 k B ⋅ T = ⟨ E ⟩ {\displaystyle {\frac {3}{2}}\,k_{\text{B}}\cdot T=\langle E\rangle }
∑ k = 1 N k 2 {\displaystyle \sum _{k=1}^{N}k^{2}}
ℏ = h 2 π {\displaystyle \hbar ={\frac {h}{2\pi }}}
Δ x ⋅ Δ p ≥ ℏ 2 {\displaystyle \Delta x\cdot \Delta p\geq {\frac {\hbar }{2}}}
λ = h p x = h 2 m E k i n {\displaystyle \lambda ={\frac {h}{p_{x}}}={\frac {h}{\sqrt {2mE_{kin}}}}}
Ψ p x ( x , t ) = A ⋅ sin ( 2 π x λ − 2 π t T ) + B ⋅ cos ( 2 π x λ − 2 π t T ) {\displaystyle \Psi _{p_{x}}(x,t)={\text{A}}\cdot {\text{sin}}\left({\frac {2\pi {\text{x}}}{\lambda }}-{\frac {2\pi {\text{t}}}{T}}\right)+{\text{B}}\cdot {\text{cos}}\left({\frac {2\pi {\text{x}}}{\lambda }}-{\frac {2\pi {\text{t}}}{T}}\right)}
Ψ E k i n ( x , t ) = A ⋅ sin ( 2 m E k i n ℏ ⋅ x − 2 π t T ) + B ⋅ sin ( 2 m E k i n ℏ ⋅ x − 2 π t T ) {\displaystyle \Psi _{E_{kin}}(x,t)={\text{A}}\cdot {\text{sin}}\left({\frac {\sqrt {2mE_{kin}}}{\hbar }}\cdot {\text{x}}-{\frac {2\pi {\text{t}}}{T}}\right)+{\text{B}}\cdot {\text{sin}}\left({\frac {\sqrt {2mE_{kin}}}{\hbar }}\cdot {\text{x}}-{\frac {2\pi {\text{t}}}{T}}\right)}
Ψ p A ( x , t ) = A ⋅ sin ( p x ℏ ⋅ x − 2 π t T ) + B ⋅ cos ( p x ℏ ⋅ x − 2 π t T ) {\displaystyle \Psi _{p_{A}}(x,t)={\text{A}}\cdot {\text{sin}}\left({\frac {p_{x}}{\hbar }}\cdot {\text{x}}-{\frac {2\pi {\text{t}}}{T}}\right)+{\text{B}}\cdot {\text{cos}}\left({\frac {p_{x}}{\hbar }}\cdot {\text{x}}-{\frac {2\pi {\text{t}}}{T}}\right)}
Ψ ′ ( x ) = d Ψ ( x ) d x {\displaystyle \Psi '(x)={\frac {d\Psi (x)}{dx}}}
− ℏ 2 2 m d 2 d x 2 Ψ E k i n ( x ) = E k i n ⋅ Ψ E k i n ( x ) {\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dx^{2}}}\Psi _{E_{k}in}(x)=E_{kin}\cdot \Psi _{E_{kin}}(x)}
E ^ K i n = − ℏ 2 2 m d 2 d x 2 {\displaystyle {\hat {E}}_{Kin}=-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dx^{2}}}}
E ^ k i n ⋅ Ψ E k i n ( x ) = E k i n ⋅ Ψ E k i n ( x ) {\displaystyle {\hat {E}}_{kin}\cdot \Psi _{E_{kin}}(x)=E_{kin}\cdot \Psi _{E_{kin}}(x)}
Fehler beim Parsen (Syntaxfehler): {\displaystyle \Psi_{Gauss}\sim\text{e} - \frac{(\text{x} - \text{x_0})^2}{2 \sigma}}
