The discussion revolves around the ground state energy of a quantum mechanical system described by a potential of the form λ²p x²/2m + 0.5kx². Participants explore the implications of the parameters involved, particularly λ and ω, and how they affect the energy calculations, including potential rescaling of mass and position.
Participants express various viewpoints on the implications of λ and the rescaling of mass and position. There is no consensus on whether the presence of λ² in the potential term would yield the same results as in the kinetic term, indicating ongoing debate and exploration of the topic.
Participants discuss the dependence on definitions of λ and ω, and the implications of rescaling mass versus position. The discussion includes unresolved mathematical steps and assumptions about the nature of the potential and kinetic energy terms.
λ is a constant. And ω=sqrt(k/m)DrClaude said:What is λ? How do you calculate ω?
Since λ is a constant, isn't it equivalent to rescaling the mass? Use that to modify the equation for ω.Apashanka das said:λ is a constant. And ω=sqrt(k/m)
sorry I didn't get youDrClaude said:Since λ is a constant, isn't it equivalent to rescaling the mass? Use that to modify the equation for ω.
But I don't see any m in the potential energy term. It is based on a spring constant k.Apashanka das said:its then kinetic energy term of mass m' but the potential energy is for mass m
ok sir in that case the ans is .5λħω or .5ħω'DrClaude said:But I don't see any m in the potential energy term. It is based on a spring constant k.
Correct.Apashanka das said:ok sir in that case the ans is .5λħω or .5ħω'
Am I right?
in that case x is replaced by x'Apashanka das said:I am having a doubt that if Id lho potential is Given of the form px2/2m+λ2.5kx2
then whether the ground state energy of the system is
λ.5ħω
No, there is no change in x. We are rescaling mass, so it will affect momentum, kinetic energy, and the frequency of oscillation, but not position.Apashanka das said:Sir once more if there
in that case x is replaced by x'
Am I right
sir what if the potential is given like px2/2m+.5λ2kx2DrClaude said:No, there is no change in x. We are rescaling mass, so it will affect momentum, kinetic energy, and the frequency of oscillation, but not position.
I think I want to let you figure this out yourself.Apashanka das said:sir what if the potential is given like px2/2m+.5λ2kx2
where is the rescaling done
sir this much part I have understand but my question is if λ2 term is present in the potential term instead of kinetic term then by rescaling x it yield the same ansDrClaude said:I think I want to let you figure this out yourself.
My approach was to try to recast the problem
$$
\hat{H} = \lambda^2 \frac{\hat{p}^2}{2m} + \frac{k}{2}x^2
$$
into one for which I know the solution (instead of having to find solutions from scratch, which is of course a viable way to do it). I see that if I set ##m' = m/\lambda^2##, then I have
$$
\hat{H} = \frac{\hat{p}^2}{2m'} + \frac{k}{2}x^2
$$
I know that the eigenenergies for this Hamiltonian are
$$
E_n = \left(n + \frac{1}{2} \right) \hbar \omega'
$$
with ##\omega' = \sqrt{k / m'}##. I also know that for ##\lambda = 1##, we recover the usual LHO with ##\omega = \sqrt{k / m}##. It is from this that I infer that the frequency of the oscillator is ##\omega' = \lambda \omega##, or
$$
E_n = \left(n + \frac{1}{2} \right) \hbar \lambda \omega
$$
So there presence of the ##\lambda^2## in the kinetic energy term is the same as changing the mass of the oscillator.
You can take a similar approach for this new potential.
I would see it more as a rescaling of the force constant k. You have to be careful because x appears also in the kinetic energy through d/dx.Apashanka das said:sir this much part I have understand but my question is if λ2 term is present in the potential term instead of kinetic term then by rescaling x it yield the same ans
Is it right