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Kinetic Energy

Oct 06, 20231 min read

Physics

Often represented with either a $K$ or $T$

Classical

$K = \frac{1}{2} m v^{2}$

$K = \frac{p ^{2}}{2 m}$

$K = \frac{3}{2} τ$

Kinetic energy of a particle
$τ$ is the fundamental temperature

Quantum

$\hat{K} = - \frac{ℏ ^{2}}{2 m} \frac{\partial ^{2}}{\partial x ^{2}}$

$⟨ K ⟩ = - \frac{ℏ ^{2}}{2 m} \nabla = \int Ψ^{*} \frac{\partial ^{2} Ψ}{\partial x ^{2}} d x$

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Classical
$\displaystyle K=\frac{1}{2}mv^{2}$
$\displaystyle K=\frac{p^{2}}{2m}$
$\displaystyle K=\frac{3}{2}\tau$
Quantum
$\displaystyle \hat{K}=- \frac{\hbar ^{2}}{2m}\frac{ \partial ^{2} }{ \partial x^{2} }$
$\displaystyle {\left\langle{K}\right\rangle}=-\frac{\hbar ^{2}}{2m}\nabla=\int \Psi^{*}\frac{ \partial ^{2}\Psi }{ \partial x^{2} } \, \mathrm{d}x$

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Kinetic Energy

Classical

$K = \frac{1}{2} m v^{2}$

$K = \frac{p ^{2}}{2 m}$

$K = \frac{3}{2} τ$

Quantum

$\hat{K} = - \frac{ℏ ^{2}}{2 m} \frac{\partial ^{2}}{\partial x ^{2}}$

$⟨ K ⟩ = - \frac{ℏ ^{2}}{2 m} \nabla = \int Ψ^{*} \frac{\partial ^{2} Ψ}{\partial x ^{2}} d x$

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Kinetic Energy

Classical

K=21​mv2

K=2mp2​

K=23​τ

Quantum

K^=−2mℏ2​∂x2∂2​

⟨K⟩=−2mℏ2​∇=∫Ψ∗∂x2∂2Ψ​dx

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$K = \frac{1}{2} m v^{2}$

$K = \frac{p ^{2}}{2 m}$

$K = \frac{3}{2} τ$

$\hat{K} = - \frac{ℏ ^{2}}{2 m} \frac{\partial ^{2}}{\partial x ^{2}}$

$⟨ K ⟩ = - \frac{ℏ ^{2}}{2 m} \nabla = \int Ψ^{*} \frac{\partial ^{2} Ψ}{\partial x ^{2}} d x$