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Show that E must be exceed the minimum value of V(x) ,for every normalizable solution to the time independent Schrodinger equation what is classical analog to this statement?

d2dx2=2mh2[V(x)E];

IfE<Vmin then and its second derivative always have the same sign. Is it normalized?

Short Answer

Expert verified

If E<Vmin, then the wave function will diverge from zero as x goes to , and therefore the wave function is not normalized, so to get a normalized wave function E>Vmin must be true.

Step by step solution

01

Step 1: Definition of Schrodinger equation 

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.

02

Identification of the equation is normalized or not

Write equation (2.5).

d2dx2=2m2[V(x)E]

If the energy E was less than the minimum value of V(x), then [VminE]>0, which is mean that the time-independent wave-function and its second derivative will have the same sign everywhere, and therefore will diverge from zero as x, which is mean is not normalized;

So that E must be greater than Vmin to have a normalized value of .

The classical analog for this is the total energy of some particle (i.e., kinetic energy plus potential energy) where it must be greater than the minimum value of the potential energy just like the energyEand the potential energyV(x)in quantum mechanics

Thus, when E<Vmin, then the wave function will diverge from zero as x goes to , and therefore the wave function is not normalized, so to get a normalized wave function E>Vmin must be true.

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Most popular questions from this chapter

Although the overall phase constant of the wave function is of no physical significance (it cancels out whenever you calculate a measurable quantity), the relative phase of the coefficients in Equation 2.17 does matter. For example, suppose we change the relative phase of 1and2in problem 2.5:(x,0)=A[1x+ei2x]Where is some constant. Find (x,t),|x,t|2, and (x), and compare your results with what you got before. Study the special cases =2and=.

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A particle in the harmonic oscillator potential starts out in the state(x,0)=A[30(x)+41(x)]

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b) Construct (x,t)and|(x,t)2|

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If two (or more) distinct44solutions to the (time-independent) Schr枚dinger equation have the same energy E . These states are said to be degenerate. For example, the free particle states are doubly degenerate-one solution representing motion to the right. And the other motion to the left. But we have never encountered normalizable degenerate solutions, and this is no accident. Prove the following theorem: In one dimension45 there are no degenerate bound states. Hint: Suppose there are two solutions, 1and 2with the same energy E. Multiply the Schr枚dinger equation for 1by 2and the Schr枚dinger equation for 2by 1and subtract, to show that 2d1/dx-2d1/dxis a constant. Use the fact that for normalizable solutions 0atto demonstrate that this constant is in fact zero.Conclude that 2s a multiple of 1and hence that the two solutions are not distinct.

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