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A particle of mass m is located in a unidimensional potential field U(x) whose shape is shown in fig. where U(0)=oo. Find (a) the equation defining the possible values of enrgy of the particle in the region E lt U_(0), reduce taht equation to the from sin kl=+- klsqrt( ħ^(2)//2ml^(2)U_(0)), where k=sqrt(2mE// ħ). Solving this equation by graphical means, domonstrate that hte possible values of energy of the particle from a discontinuous spectrum, (b)the minium value of the quantity l^(2)U_(0) at which the first energy level appears in the region E lt U_(0). At what minimum value of l^(2)U_(0) does the nth level appear? |
Answer» SOLUTION : (a) Starting from the Schrodinger equation in the regions I &II `(d^(2)Psi)/(dx^(2))+(2mE)/( ħ^(2))Psi=0 xi n I`....(1) `(d^(2)Psi)/(dx^(2))-(2mE(U_(0)-E))/ ħ^(2)Psi=0x` in II....(2) where `U_(0) gt E gt 0`, we easily derive the solutions in I &II (3) `Psi_(1)(x)=A sin kx+Bcoskx`....(3) `Psi_(n)(x)=ce^(alphax)+De^(-ax)`....(4) where `k^(2)=(2mE)/( ħ^(2)),alpha^(2)=(2m(U_(0)-E))/(ħ^(2))` The boundary conditions are `Psi(o)=0`....(5) and `Psi &((dPsi)/(dx))` are continous at `x=l`, and `Psi` MUST vanish at `x=+oo` Then `Psi_(I)= A sin kx` and `Psi_(II)=De^(-alpha x)` So `A sin kl=De^(-alpha l)` `kA cos kl= -alpha De^(-alphal)` Frim this we get `tan kl=-(k)/(alpha)` or `sin kl= +-kl//sqrt(k^(2)l^(2)+alpha^(2)l^(2))` `=+-kl//sqrt((2mU_(0)l^(2))/(ħ^(2)))` `=+-kl sqrt(ħ^(2)//2mU_(0)l^(2))` ....(6)  Plotting the left and right siders of this equation we can find the points at which the straight lines cross the sine curve. The roots of the equation corresponding to the eign values of energy `E_(i)` and FOUND from the intersection point `(kl)_(i)`, for which tan `(kl)_(i)lt 0`(i.e., `2^(nd)& 4^(th)` and other even quadrants). It seen that bound states do not always exist. Find the first bound state to appear (refer to the line (B) above) `(kl)_(1,mi n)=(pi)/(2)` (b) Substituting, we get `(l^(2)U_(0))_(1,mi n)=(pi^(2)h^(2))/(8m)` as the conditions for the apperance of the first bound state. The second bound state will appear when `kl` is in fourth quadrant. The magnitude of the slope of the straight line must then be less than `(1)/(3pi//2)` Corresponding `(kl)_(2,mi n)=(3pi)/(2)=(3)(pi)/(2)=(2xx2-1)(pi)/(2)` For `n` bound state, it is easy to convince one self that the slope of the APPROPRIATE straight line (upper or lower) must be less than `(kl)_(n,mi n)=(2n-1)(pi)/(2)` Then `(t^(2)U_(0))_(n,mi n)=((2n-1)^(2)pi^(2) ħ^(2))/(8m)` Do not forget to note that for large `n` both+and -signs in the Eq.(6) contribute to solutions. |
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