It is not easy to find the solution of Schroeder's equation for most realistic potential field. The simplest example is

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It Is Not Easy To Find The Solution Of Schroeder S Equation For Most Realistic Potential Field The Simplest Example Is 1 (79.74 KiB) Viewed 33 times

It is not easy to find the solution of Schroeder's equation for most realistic potential field. The simplest example is the question of particles in a potential energy well with infinite boundary values as shown in the V(X) 0 L . figure. It is assumed that the particle is V(x)=0 except for the boundaries x=0 and x=L, and V(x) is trapped in an infinitely large potential well at the boundary. (x,t)= Ae-j/)(Et-px) V(x) = 0 . 0<x<L ; V(x) = 0, x=0. The exact expression of the wave function, which is a solution obtained by solving the one-dimension Schoedunger equation independent of time inside the well after separating the given wave function by variables, is as follows. 117 V = 2 L Suppose a particle lies in the ground state of the potential energy well given above. (a) In the ground state, we want to find the expectation for standard deviation (uncertainty Ax in the position of the particle), Ax of the measurement value for the position. Using the given conditions and wave functions mentioned above, describe the expected value of Ax in the expression of L(calculation process required) (b)in the ground state, we want to find the expectations for the standard deviation (uncertainty 4px, in the momentum of the particle). Apx of the measurement for momentum. Using the given conditions and wave functions mentioned above, describe the expected value of Apx in the expression of L (calculation process required) (c)Using the Ax and Apx values obtained above, induce Heisenberg's Uncertainty Principle to be Ax4px>=ħ/2.(calculation process required)