Quantum Hamilton-Jacobi Solutions for 1-Dimensional by S. ranjani

By S. ranjani

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10), implies that we select the value of b′1 , which is positive in the limit h ¯ → 0. 23) 3¯ h 4 if (A + B) < 0. From the above equation, we see that both the values of b′1 are acceptable, but in two different ranges. 3), show that they correspond to the different phases of SUSY. 24) which is positive. The two values of the residue of χ are b1 = ¯ (A − B) h + , 2α 4 − (A − B) 3¯ h + . 26) Chapter 3. Study of potentials exhibiting different spectra 43 But, the present case of interest is (A − B) > 0, for which the choice of residue is b1 = ¯ (A − B) h + .

Study of potentials exhibiting different spectra 51 functions of one dimensional potentials are non- degenerate, which implies, the wave functions are real. Hence, χ must be real, for which we must have b1 = b′1 . 48) becomes 2b1 + n = d1 . 58) Thus, we see that the property, the wave function should be finite for x → ∞, gives the values of d1 in the two ranges as follows d1 =      1±2s 2 for 0 < s < 1/2 1−2s 2 for s > 1/2. 59) From the parity constraint, one obtains the restriction on the values of the residues at the fixed poles as b1 = b′1 .

This is equivalent to what one gets from the exact quantization condition and is widely used in our analysis of periodic and PT symmetric potentials. 36) are two crucial steps in arriving at the correct physical solutions. For many problems, it is necessary to perform a change of variable y = f (x). 67) 31 Chapter 2. 69) expressed as a function of y and V˜ (y) = V (x(y)). 69) to a simple form, by introducing χ through the following transformation equations. 71) where, the residue at moving poles is one.

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