91影视

A differential equation that 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.

The boundary conditions

$\mathrm{蠄}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}{\mathrm{Ae}}^{\mathrm{ikx}}+{\mathrm{Be}}^{鈭�\mathrm{ikx}}& (\mathrm{x}<鈭�\mathrm{a})\\ {\mathrm{Ce}}^{\mathrm{ikx}}+{\mathrm{De}}^{鈭�\mathrm{ikx}}& (鈭�\mathrm{a}<\mathrm{x}<\mathrm{a})\\ {\mathrm{Fe}}^{\mathrm{ikx}}& (\mathrm{a}>\mathrm{x})\end{array}\right.$

Using the continuity at$-a$

$A{e}^{ika}+B{e}^{鈭�ika}=C{e}^{鈭�ika}+D{e}^{ika}$

Let $尾=e^{鈭�2ika}$

So $尾A+B=尾C+D$ 鈥�(颈)

Using the continuity at$+a$

$C{e}^{ika}+D{e}^{鈭�ika}=F{e}^{ika}$

So $F=C+尾D$ 鈥�(颈颈)

Using the discontinuity of ${蠄}^{\text{'}}$at$-a$

$ik(C{e}^{鈭�ika}鈭�D{e}^{ika})鈭�ik(A{e}^{鈭�ika}鈭�B{e}^{ika})=\frac{鈭�2m伪}{{\mathrm{鈩�}}^2}(A{e}^{鈭�ika}+B{e}^{ika})$

Let $纬=i2m伪/{\mathrm{鈩�}}^{2}k$

So $尾C鈭�D=尾(纬+1)A+B(纬鈭�1)$ 鈥� (iii)

Using the discontinuity of ${蠄}^{\text{'}}$at$+a$

$ikF{e}^{ika}鈭�ik(C{e}^{ika}鈭�D{e}^{鈭�ika})=\frac{鈭�2m伪}{{\mathrm{鈩�}}^2}\left(F{e}^{ika}\right)$

So $C鈭�尾D=(1鈭�纬)F$ 鈥�(颈惫)

Adding (2) and (4)

$2C=F+(1鈭�纬)F$

So $2C=(2鈭�纬)F$

Subtract (ii) and (iv)

$2尾D=F鈭�(1鈭�纬)F$

so , $2D=(纬/尾)F$

Add (i) and (iii)

$2尾C=尾A+B+尾(纬+1)A+B(纬鈭�1)$

So$2C=(纬+2)A+(纬/尾)B.$

Equation 2C in the equations

$(2鈭�纬)F=(纬+2)A+(纬/尾)B$ 鈥�(惫)

Equation 2D in the equations

$(纬/尾)F=鈭�纬尾A+(2鈭�纬)B$ 鈥�(惫颈)

$尾{(2鈭�纬)}^{2}F=尾(4鈭�{纬}^2)A+纬(2鈭�纬)B$

$({纬}^2/尾)F=鈭�尾{纬}^{2}A+纬(2鈭�纬)B$

$[尾{(2鈭�纬)}^2鈭�{纬}^2/尾]F=尾[4鈭�{纬}^2+{纬}^2]A=4尾A$

So $\frac{\mathrm{F}}{\mathrm{A}}=\frac{4}{(2鈭�{\mathrm{纬}}^2)鈭�{\mathrm{纬}}^2/{\mathrm{尾}}^2}$

Let $g=i/纬=\frac{{\mathrm{鈩�}}^{2}k}{2m伪}$ and $蠒=4kA$

So$纬=\frac{i}{g},\text{鈥夆赌夆赌�}{尾}^2=e^{鈭�i蠒}$

$\frac{F}{A}=\frac{4g^2}{{(2g鈭�i)}^2+e^{i蠒}}$

The Denominator: $4g^2鈭�4ig鈭�1+\cos 蠒+i\sin 蠒=(4g^2鈭�1+\cos 蠒)+i(\sin 蠒鈭�4g)$

${\left[\text{TheDenominator}\right]}^2={(4g^2鈭�1+\cos 蠒)}^2鈭�{(\sin 蠒鈭�4g)}^2$

$=16g^4+1+{\cos }^2蠒鈭�8g^2鈭�2\cos 蠒+8g^2\cos 蠒+\sin 2蠒鈭�8g\sin 蠒+16g^2$

$=16g^4+8g^2+2+(8g^2鈭�2)\cos 蠒鈭�8g\sin 蠒.$

Therefore, the transmission coefficient for the potential is

$T={\left|\frac{F}{A}\right|}^2=\frac{8g^4}{(8g^4+4g^2+1)+(4g^2鈭�1)\cos 蠒鈭�4g\sin 蠒}$.