InterviewSolution
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InterviewSolution
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Using the Schrodinger equation, demonstrate that at the point where the potential energy `U(x)` of a particle has a infinte discountinuity, the wave function renains smooth, I.e., its first derivaltive with respect to the coordinate is continuous. |
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Answer» We can for definiteness assume that the discontinuity occurs at the point `x=0`. Now the schordinger equation is `( ħ^(2))/(2m)(d^(2)Psi)/(dx^(2))+U(x)Psi(x)=EPsi(x)` We intergate this equation around `x=0`i.e., from `x=-epsilon_(1) to x=epsilon_(2)` where `epsilon_(1),epsilon_(2)` are small positives numbers. Then `( ħ^(2))/(2m)int_(-epsilon_(1))^(+epsilon_(2))(d^(2)Psi)/(dx^(2))dx=int_(-epsilon_(1))^(+epsilon_(2))(E-U(x)Psi(x)dx)` ltbrltgt or `((dPsi)/(dx))_(+epsilon_(2))=((d Psi)/(dx))_(-epsilon_(1))=-(2m)/ (ħ^(2)) int_(-epsilon_(1))^(epsilon_(2))(E-U(x))_(dx)Psi(x)` SInce the potential and the energy `E` are finite and `Psi(x)` is bounded by assumption, the intergaral on the right exists and `rarr0 asepsilon_(1),epsilon_(2)rarr0` Thus `((d Psi)/(dx))_(epsilon_(2))=((dPsi)/(dx))_(-e_(1)) as epsilon_(1),epsilon_(2)rarr0` So `((dPsi)/(dx))` is continuous at `x=0` (the point where `U(x)` has a infinite jump discontuinuityI `((d Psi)/(dx))_(epsilon_(2))=((dPsi)/(dx))_(-e_(1)) as epsilon_(1),epsilon_(2)rarr0` So `((dPsi)/(dx))` is continuous at `x=0` (the point where `U(x)` has a infinite jump discontuinuity. |
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