Ground State Energy of a Potential: λ2px2/2m+0.5kx2

[Apashanka das]

I am having a doubt that if Id lho potential is Given of the form λ²p_x²/2m+.5kx²
then whether the ground state energy of the system is
λ²/4ħω+1/4ħω

 

[DrClaude]

What is λ? How do you calculate ω?

 

[Apashanka das]

What is λ? How do you calculate ω?

λ is a constant. And ω=sqrt(k/m)

 

[DrClaude]

λ is a constant. And ω=sqrt(k/m)

Since λ is a constant, isn't it equivalent to rescaling the mass? Use that to modify the equation for ω.

 

[Apashanka das]

Since λ is a constant, isn't it equivalent to rescaling the mass? Use that to modify the equation for ω.

sorry I didn't get you

 

[DrClaude]

Set ##m' = m / \lambda^2##. What do you get then?

 

[DrClaude]

its then kinetic energy term of mass m' but the potential energy is for mass m

But I don't see any m in the potential energy term. It is based on a spring constant k.

 

[Apashanka das]

But I don't see any m in the potential energy term. It is based on a spring constant k.

ok sir in that case the ans is .5λħω or .5ħω'
Am I right?
Thanks sir in helping me to understand the concept

 

[DrClaude]

ok sir in that case the ans is .5λħω or .5ħω'
Am I right?

Correct.

 

[Apashanka das]

Sir once more if there

I am having a doubt that if Id lho potential is Given of the form p_x²/2m+λ².5kx²
then whether the ground state energy of the system is
λ.5ħω

in that case x is replaced by x'
Am I right

 

[DrClaude]

Sir once more if there

in that case x is replaced by x'
Am I right

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]

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.

sir what if the potential is given like px²/2m+.5λ²kx²
where is the rescaling done

 

[DrClaude]

sir what if the potential is given like px²/2m+.5λ²kx²
where is the rescaling done

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.

 

[Apashanka das]

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.

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

 

[DrClaude]

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

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.