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A differential equation describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.

_{$V\left(x\right)=0$} When _{localid="1656057126624" $x鈮�鈭�a$}

_{$V\left(x\right)=鈭�V_0$} When _{$鈭�\mathrm{a}<\mathrm{x}<\mathrm{a}$}

_{$V\left(x\right)=0$} When _{$x鈮�鈭�a$}

Schrodinger equation

$鈭�\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\mathrm{唯}\left(\mathrm{x}\right)+\mathrm{V}\left(\mathrm{x}\right)\mathrm{唯}\left(\mathrm{x}\right)=\mathrm{贰唯}\left(\mathrm{x}\right)$

We shall split into 3 regions

$\mathrm{唯}\left(\mathrm{x}\right)={\mathrm{Ae}}^{\mathrm{ikx}}+{\mathrm{Be}}^{鈭�\mathrm{ikx}}$ $\mathrm{x}<鈭�\mathrm{a}$

$\mathrm{唯}\left(\mathrm{x}\right)={\mathrm{Fe}}^{\mathrm{ikx}}$$\mathrm{x}>\mathrm{a}$

${\mathrm{Ae}}^{鈭�\mathrm{ika}}+{\mathrm{Be}}^{\mathrm{ika}}={\mathrm{Ce}}^{鈭�\mathrm{渭补}}+{\mathrm{De}}^{\mathrm{渭补}}$

$\mathrm{ik}({{\mathrm{Ae}}^{鈭�}}^{\mathrm{ika}}鈭�{\mathrm{Be}}^{\mathrm{ika}})=\mathrm{渭}({\mathrm{Ce}}^{鈭�\mathrm{渭补}}鈭�{\mathrm{De}}^{\mathrm{渭补}})$

We have

_{${\mathrm{Ce}}^{\mathrm{渭补}}+{\mathrm{De}}^{\mathrm{渭补}}={\mathrm{Fe}}^{\mathrm{ika}}$}

_{${\mathrm{ikFe}}^{\mathrm{ika}}=\mathrm{渭}({\mathrm{Ce}}^{鈭�\mathrm{渭补}}鈭�{\mathrm{De}}^{\mathrm{渭补}})$}

$\mathrm{B}=\frac{{\mathrm{e}}^{鈭�2\mathrm{ika}}({\mathrm{k}}^2+{\mathrm{渭}}^2)\sinh \left(2\mathrm{渭补}\right)}{2\mathrm{颈办渭肠辞蝉丑}\left(2\mathrm{渭补}\right)+({\mathrm{k}}^2鈭�{\mathrm{渭}}^2)\sinh \left(2\mathrm{渭补}\right)}\mathrm{A}$

$\mathrm{C}=\frac{{{\mathrm{e}}^{鈭�\mathrm{补渭}}}^{鈭�\mathrm{ika}}(鈭�{\mathrm{k}}^2+\mathrm{办渭颈})}{2\mathrm{颈办渭肠辞蝉丑}\left(2\mathrm{渭补}\right)+({\mathrm{k}}^2鈭�{\mathrm{渭}}^2)\sinh \left(2\mathrm{渭补}\right)}\mathrm{A}$

$\mathrm{D}=\frac{{{\mathrm{e}}^{鈭�\mathrm{补渭}}}^{鈭�\mathrm{ika}}(鈭�{\mathrm{k}}^2+\mathrm{办渭颈})}{2\mathrm{颈办渭肠辞蝉丑}\left(2\mathrm{渭补}\right)+({\mathrm{k}}^2鈭�{\mathrm{渭}}^2)\sinh \left(2\mathrm{渭补}\right)}\mathrm{A}$

$\mathrm{F}=\frac{2{\mathrm{e}}^{鈭�2\mathrm{ika}}\mathrm{办渭颈}}{2\mathrm{颈办渭肠辞蝉丑}\left(2\mathrm{渭补}\right)+({\mathrm{k}}^2鈭�{\mathrm{渭}}^2)\sinh \left(2\mathrm{渭补}\right)}\mathrm{A}$

$T={\left(\frac{F}{A}\right)}^2$

$\mathrm{T}={\left[\frac{2{\mathrm{e}}^{鈭�2\mathrm{ika}}\mathrm{办渭颈}}{2\mathrm{颈办渭肠辞蝉丑}\left(2\mathrm{渭补}\right)+({\mathrm{k}}^2鈭�{\mathrm{渭}}^2)\sinh \left(2\mathrm{渭补}\right)}\right]}^2$

$\mathrm{T}=\frac{4{\mathrm{渭}}^2{\mathrm{k}}^2}{({\mathrm{渭}}^4+2{\mathrm{渭}}^2{\mathrm{k}}^2+{\mathrm{k}}^4)\left[{\sinh }^2\right(2\mathrm{渭补})+4{\mathrm{渭}}^2{\mathrm{k}}^2{\cosh }^2(2\mathrm{渭补}\left)\right]}$

$\mathrm{Put}{\cosh }^2\left(2\mathrm{渭补}\right)=1+{\sinh }^2\left(2\mathrm{渭补}\right)$

$\mathrm{T}=\frac{1}{{({\mathrm{渭}}^2+{\mathrm{k}}^2)}^2{\sinh }^2\left(2\mathrm{渭补}\right)/4{\mathrm{渭}}^2{\mathrm{k}}^2+1}$

${\mathrm{T}}^{鈭�1}=1+\frac{({\mathrm{渭}}^2+{\mathrm{k}}^2){\sinh }^2\left(2\mathrm{渭补}\right)}{4{\mathrm{渭}}^2{\mathrm{k}}^2}$

Put the value of constant$\mathrm{渭}$and$\mathrm{k}$we get,

${\mathrm{T}}^{鈭�1}=1+\frac{{{\mathrm{V}}_0}^2}{4\mathrm{E}({\mathrm{V}}_0鈭�\mathrm{E})}{\sinh }^2\left(\frac{2\mathrm{a}}{\mathrm{h}}\sqrt{2\mathrm{m}({\mathrm{V}}_0鈭�\mathrm{E})}\right)$

When energy is equal to the potential but the barrier region we have

${\mathrm{唯}}^{\mathrm{n}}=0$

$\mathrm{唯}\left(\mathrm{x}\right)=\mathrm{Cx}+\mathrm{D}$

${\mathrm{Ae}}^{鈭�\mathrm{ika}}+{\mathrm{Be}}^{\mathrm{ika}}=鈭�\mathrm{Ca}+\mathrm{D}$

$\mathrm{ik}({\mathrm{Ae}}^{鈭�\mathrm{ika}}鈭�{\mathrm{Be}}^{\mathrm{ika}})=\mathrm{C}$

At, $\mathrm{x}=\mathrm{a}$

$\mathrm{Ca}+\mathrm{D}={\mathrm{Fe}}^{\mathrm{ika}}$

$\mathrm{C}={\mathrm{ikFe}}^{\mathrm{ika}}$

$\mathrm{B}=\left(\frac{{\mathrm{kae}}^{鈭�2\mathrm{ika}}}{\mathrm{ka}+\mathrm{i}}\right)\mathrm{A}$

$\mathrm{C}=\left(\frac{{\mathrm{ke}}^{鈭�\mathrm{ika}}}{鈭�\mathrm{ka}鈭�\mathrm{i}}\right)\mathrm{A}$

$\mathrm{D}={\mathrm{e}}^{鈭�\mathrm{ika}}\mathrm{A}$

$\mathrm{F}=\left(\frac{{\mathrm{e}}^{鈭�2\mathrm{ika}}}{1鈭�\mathrm{ika}}\right)\mathrm{A}$

$\mathrm{T}={\left(\frac{\mathrm{F}}{\mathrm{A}}\right)}^2$

$\mathrm{T}=\frac{1}{1+\frac{2{\mathrm{mEa}}^2}{{\mathrm{h}}^2}}$

${\mathrm{T}}^{鈭�1}=1+\frac{2{\mathrm{mEa}}^2}{{\mathrm{h}}^2}$

$鈭�\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\mathrm{唯}\left(\mathrm{x}\right)+\mathrm{V}\left(\mathrm{x}\right)\mathrm{唯}\left(\mathrm{x}\right)=\mathrm{贰唯}\left(\mathrm{x}\right)$

$鈭�\frac{{\mathrm{h}}^2}{2\mathrm{m}}{\mathrm{唯}}^{\mathrm{n}}=(\mathrm{E}鈭�{\mathrm{V}}_0)\mathrm{唯}$

${\mathrm{唯}}^{\mathrm{n}}=鈭�\frac{2\mathrm{m}(\mathrm{E}鈭�{\mathrm{V}}_0)}{{\mathrm{h}}^2}$

${\mathrm{唯}}^{\mathrm{n}}=鈭�{\mathrm{位}}^2\mathrm{唯}$

$\mathrm{位}=\sqrt{\frac{2\mathrm{m}(\mathrm{E}鈭�{\mathrm{V}}_0)}{{\mathrm{h}}^2}}$

${\mathrm{Ae}}^{鈭�\mathrm{ika}}+{\mathrm{Be}}^{\mathrm{ika}}={\mathrm{Ce}}^{鈭�\mathrm{颈位补}}+{\mathrm{De}}^{\mathrm{颈位补}}$

$\mathrm{ik}({{\mathrm{Ae}}^{鈭�}}^{\mathrm{ika}}鈭�{\mathrm{Be}}^{\mathrm{ika}})=\mathrm{颈位}({\mathrm{Ce}}^{鈭�\mathrm{颈位补}}鈭�{\mathrm{De}}^{\mathrm{颈位补}})$

We have

${\mathrm{Ce}}^{\mathrm{颈位补}}+{{\mathrm{De}}^{鈭�}}^{\mathrm{颈位补}}={\mathrm{Fe}}^{\mathrm{ika}}$

${\mathrm{ikFe}}^{\mathrm{ika}}=\mathrm{颈位}({\mathrm{Ce}}^{鈭�\mathrm{颈位补}}鈭�{\mathrm{De}}^{\mathrm{颈位补}})$

$\mathrm{B}=\frac{{\mathrm{e}}^{鈭�2\mathrm{ika}}({\mathrm{k}}^2鈭�{\mathrm{位}}^2)\sin \left(2\mathrm{位补}\right)}{2\mathrm{办位肠辞蝉}\left(2\mathrm{位补}\right)+({\mathrm{k}}^2+{\mathrm{位}}^2)\sin \left(2\mathrm{位补}\right)}\mathrm{A}$

$\mathrm{C}=\frac{{\mathrm{e}}^{鈭�\mathrm{ia}(\mathrm{位}+\mathrm{k})}(\mathrm{位}+\mathrm{K})\mathrm{K}}{2\mathrm{办位肠辞蝉}\left(2\mathrm{位补}\right)鈭�\mathrm{i}({\mathrm{k}}^2+{\mathrm{位}}^2)\sin \left(2\mathrm{位补}\right)}\mathrm{A}$

$\mathrm{D}=\frac{{\mathrm{e}}^{鈭�\mathrm{ia}(\mathrm{k}鈭�\mathrm{位})}(\mathrm{k}鈭�\mathrm{位})\mathrm{K}}{鈭�2\mathrm{办位肠辞蝉}\left(2\mathrm{位补}\right)+\mathrm{i}({\mathrm{k}}^2+{\mathrm{位}}^2)\sin \left(2\mathrm{位补}\right)}\mathrm{A}$

$\mathrm{F}=\frac{2{\mathrm{办位别}}^{鈭�2\mathrm{ika}}}{2\mathrm{办位肠辞蝉}\left(2\mathrm{位补}\right)鈭�\mathrm{i}({\mathrm{k}}^2+{\mathrm{位}}^2)\sin \left(2\mathrm{位补}\right)}\mathrm{A}$

$\mathrm{T}={\left(\frac{\mathrm{F}}{\mathrm{A}}\right)}^2={\left(\frac{2{\mathrm{办位别}}^{鈭�2\mathrm{ika}}}{2\mathrm{办位肠辞蝉}\left(2\mathrm{位补}\right)鈭�\mathrm{i}({\mathrm{k}}^2+{\mathrm{位}}^2)\sin \left(2\mathrm{位补}\right)}\right)}^2$

$\mathrm{T}=\frac{4{\mathrm{位}}^2{\mathrm{k}}^2}{({\mathrm{位}}^4鈭�2{\mathrm{位}}^2{\mathrm{k}}^2+{\mathrm{k}}^4){\sin }^2\left(2\mathrm{位补}\right)+4{\mathrm{位}}^2{\mathrm{k}}^2}$

$\mathrm{T}=\frac{1}{1+({\mathrm{位}}^2鈭�{\mathrm{k}}^2){\sin }^2\left(2\mathrm{位补}\right)/4{\mathrm{位}}^2{\mathrm{k}}^2}$

${\mathrm{T}}^{鈭�1}=1+\frac{({\mathrm{位}}^2鈭�{\mathrm{k}}^2){\sin }^2\left(2\mathrm{位补}\right)}{4{\mathrm{位}}^2{\mathrm{k}}^2}$

${\mathrm{T}}^{鈭�1}=1+\frac{{{\mathrm{V}}_0}^2}{4\mathrm{E}({\mathrm{V}}_0鈭�\mathrm{E})}{\sin }^2\left(\frac{2\mathrm{a}}{\mathrm{h}}\sqrt{2\mathrm{m}({\mathrm{V}}_0鈭�\mathrm{E})}\right)$