- #1

Baibhab Bose

- 34

- 3

- Homework Statement
- A particle in a one-dimensional harmonic oscillator potential is described by a wave-function Psi(x,t)

߰ If the wavefunction changes to ߰Psi(Lambda*x ,t) the expectation value of kinetic energy ܶ

and the potential energy ܸ will change, respectively, to

- Relevant Equations
- The Schrodinger equation

The wavefunction is Ψ(x,t) ----> Ψ(λx,t)

What are the effects on <T> (av Kinetic energy) and V (potential energy) in terms of λ?

From ## \frac {h^2}{2m} \frac {\partial^2\psi(x,t)}{\partial x^2} + V(x,t)\psi(x,t)=E\psi(x,t) ##

if we replace x by ## \lambda x ## then it becomes ## \frac {h^2}{2m} \frac {\partial^2\psi(\lambda x,t)}{\partial x^2}\frac {1}{\lambda^2} + \lambda^2 V(\lambda x,t)\psi(\lambda x,t)=E\psi(\lambda x,t) ##

so would the argument be like, to keep the energy as it is the first term should be neutralized by and extra ##\lambda^2## term in the average KE (interpretation of the first term). and similarly V(##\lambda x ,t)## should be divided by ##\lambda^2## ?

and again, why should I assume that the total energy remains same?

What are the effects on <T> (av Kinetic energy) and V (potential energy) in terms of λ?

From ## \frac {h^2}{2m} \frac {\partial^2\psi(x,t)}{\partial x^2} + V(x,t)\psi(x,t)=E\psi(x,t) ##

if we replace x by ## \lambda x ## then it becomes ## \frac {h^2}{2m} \frac {\partial^2\psi(\lambda x,t)}{\partial x^2}\frac {1}{\lambda^2} + \lambda^2 V(\lambda x,t)\psi(\lambda x,t)=E\psi(\lambda x,t) ##

so would the argument be like, to keep the energy as it is the first term should be neutralized by and extra ##\lambda^2## term in the average KE (interpretation of the first term). and similarly V(##\lambda x ,t)## should be divided by ##\lambda^2## ?

and again, why should I assume that the total energy remains same?