# Effects of KE & PE of a Harmonic Oscillator under Re-scaling of coordinates

• Baibhab Bose
In summary, when the wavefunction is changed from Ψ(x,t) to Ψ(λx,t), the kinetic energy and potential energy are affected in terms of λ. The time-independent Schrodinger equation is used to determine the energy eigenvalue, and by substituting u=λx, the equation can be rewritten in terms of u. The correct form of the equation is -h^2/2m * d^2 Ψ(λx)/dx^2 + (1/2) mω^2x^2Ψ(λx) = EΨ(λx). The kinetic energy is then given by <T> = (h^2/2m) * ∫ Ψ
Baibhab Bose
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?

Baibhab Bose said:
From ## \frac {h^2}{2m} \frac {\partial^2\psi(x,t)}{\partial x^2} + V(x,t)\psi(x,t)=E\psi(x,t) ##
The above is not a correct form for the Schrodinger equation either time-dependent or time-independent. I suggest that you stick with the time-independent form.
Avoid guessing. What if you made the substitution ##u = \lambda x##? Then you should be able to cast the TISE in terms of ##u## and deduce the energy by comparison with the usual form when ##\lambda =1##.

Yes, so then if I write KE in terms of u, i.e. ## \frac{ħ^2}{2m}\frac {\partial^2 \psi (u)}{\partial u^2}=<T>##
and multiplying by ##\lambda^2## both sides ##\lambda^2\frac{ħ^2}{2m}\frac{1}{\lambda^2}\frac {\partial^2 \psi (\lambda x)}{\partial x^2}=\lambda^2<T>##
=>##\frac{ħ^2}{2m}\frac {\partial^2 \psi (\lambda x)}{\partial x^2}=\lambda^2<T>##
this is how we arrive at the conclusion right?

Baibhab Bose said:
Yes, so then if I write KE in terms of u, i.e. ## \frac{ħ^2}{2m}\frac {\partial^2 \psi (u)}{\partial u^2}=<T>##
and multiplying by ##\lambda^2## both sides ##\lambda^2\frac{ħ^2}{2m}\frac{1}{\lambda^2}\frac {\partial^2 \psi (\lambda x)}{\partial x^2}=\lambda^2<T>##
=>##\frac{ħ^2}{2m}\frac {\partial^2 \psi (\lambda x)}{\partial x^2}=\lambda^2<T>##
this is how we arrive at the conclusion right?
¥ou did not write the time-indepemdemt Schrodinger equation in terms of ##u##, you wrote what you think is the expectation value for the kinetic energy, which it isn't. On the left side you have a function and on the right side you have a constant. It looks like you are confused about quantum mechanical operators and expectation values.

So ##\frac{ħ^2}{2m}\frac{\partial^2\psi(u)}{\partial u^2}+V(u)\psi(u)=E\psi(u)## is the time independent Schrodinger equation in terms of ## u=\lambda x##... (1)
so this E is equivalent to the case when ##\frac{ħ^2}{2m}\frac{\partial^2\psi(x)}{\partial x^2}+V(x)\psi(x)=E\psi(x)##...(2)
the same energy eigenvalue.
Now when we express the (1) expanding u as ##\lambda x## then it becomes
##\frac{ħ^2}{2m}\frac{\partial^2\psi(\lambda x)}{\partial x^2}\frac{1}{\lambda^2}+\lambda^2V(x)\psi(\lambda x)=E\psi(\lambda x)## since Harmonic oscillator potential is ##1/2mω^2 x^2##.
Now how can I draw conclusions about Kinetic and Potential energies? How they have changed due to this scale shift..?

First off there is a negative sign in front of the ##\frac{\hbar^2}{2m}## term on the Schrodinger equation. Secondly, you still don't get it. Start with$$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi(x)}{\partial x^2}+\frac{1}{2}m\omega^2 x^2\psi(x)=E\psi(x)$$ and then let ##u=\lambda~x##. Write the equation above in terms of ##u## and then bring it to a form so that the kinetic energy term is ##-\dfrac{\hbar^2}{2m}\dfrac{\partial^2 \psi(u)}{\partial u^2}.## Compare what you get with the equation above.

Okay, if I substitute ##u=\lambda x## to the ## -\frac{ħ^2}{2m} \frac{\partial^2\psi(x)}{\partial x^2}+V(x)\psi(x)=E\psi(x)## , then it becomes something like
## -\frac{ħ^2}{2m} \frac{\partial^2\psi(\frac{u}{\lambda})}{\partial u^2}\lambda^2+V(u)\psi(\frac{u}{\lambda})\frac{1}{\lambda^2}=E\psi(\frac{u}{\lambda})##
now what can I do to make this ##\psi(\frac{u}{\lambda}) ## to
##\psi(u)## ?

OK, let's do this a bit differently. In post #1you had the right idea. If the wavefunction changes from ##\psi(x)## to ##\psi(\lambda x)## as the problem suggests, the correct time-independent equation is
$$-\frac {h^2}{2m} \frac {d^2\psi(\lambda x)}{d x^2} + \frac{1}{2} m\omega^2x^2\psi(\lambda x)=E\psi(\lambda x)$$which is not what you have. As I already suggested in post #6, write the equation above in terms of ##u## and then bring it to a form so that the kinetic energy term is ##-\dfrac{\hbar^2}{2m}\dfrac{d^2 \psi(u)}{d u^2}## and there are no ##x## terms in the equation but only ##u## terms.

kuruman said:
OK, let's do this a bit differently. In post #1you had the right idea. If the wavefunction changes from ##\psi(x)## to ##\psi(\lambda x)## as the problem suggests, the correct time-independent equation is
$$-\frac {h^2}{2m} \frac {d^2\psi(\lambda x)}{d x^2} + \frac{1}{2} m\omega^2x^2\psi(\lambda x)=E\psi(\lambda x)$$which is not what you have. As I already suggested in post #6, write the equation above in terms of ##u## and then bring it to a form so that the kinetic energy term is ##-\dfrac{\hbar^2}{2m}\dfrac{d^2 \psi(u)}{d u^2}## and there are no ##x## terms in the equation but only ##u## terms.
Yes. this is clear now. I am getting the right answer.
By I want to discuss and alternate with you.
Syntactically ##<T>=\frac{h^2}{2m}\int \psi(x)* \frac{\partial^2}{\partial x^2} \psi(x)\, dx##
So, here if we change ##\psi(x)--> \psi(\lambda x)## then to scale we need to shoot two lambdas down the partial term and one for the dx term to make it fully in terms of##u=\lambda x## but, consequently there will be two lambdas in the numerator and one in denominator; effectively one lambda term for this whole kinetic energy.
So, the KE re-scale amounts to ##\lambda <T>##
If this is true, it'd be weird because none of the options in my question mentions this kind of answer.
but the logic is air, don't you think?

When I wrote
kuruman said:
As I already suggested in post #6, write the equation above in terms of ##u## and then bring it to a form so that the kinetic energy term is ##-\dfrac{\hbar^2}{2m}\dfrac{d^2 \psi(u)}{d u^2}## and there are no ##x## terms in the equation but only ##u## terms.
I meant write the entire Schrodinger equation, not what you think is the expectation value of the kinetic energy. Do that and the logic will not appear to be air. Also, if you have to "shoot lambdas", do not describe it with words, but show it algebraically with equations. It helps spotting possible mistakes.

Baibhab Bose said:
By I want to discuss and alternate with you.
Syntactically ##<T>=-\frac{h^2}{2m}\int \psi(x)* \frac{\partial^2}{\partial x^2} \psi(x)\, dx##
So, here if we change ##\psi(x) \to \psi(\lambda x)## then to scale we need to shoot two lambdas down the partial term and one for the dx term to make it fully in terms of##u=\lambda x## but, consequently there will be two lambdas in the numerator and one in denominator; effectively one lambda term for this whole kinetic energy, so the KE re-scale amounts to ##\lambda <T>##.
Your reasoning is right here, but you still need to account for the fact that ##\psi(\lambda x)## is not normalized.

I'm wondering if you provided the problem statement exactly as it was given to you. @kuruman's discussing what happens when you transform the coordinate ##x \to \lambda x## whereas the problem statement seems to asking a different question where you replace ##\psi(x)## by a new state ##\psi(\lambda x)##. In the latter case, if ##\psi(x)## is a solution to the TISE, ##\psi(\lambda x)## won't be one, but you can still calculate expectation values.

vela said:
Your reasoning is right here, but you still need to account for the fact that ##\psi(\lambda x)## is not normalized.

I'm wondering if you provided the problem statement exactly as it was given to you. @kuruman's discussing what happens when you transform the coordinate ##x \to \lambda x## whereas the problem statement seems to asking a different question where you replace ##\psi(x)## by a new state ##\psi(\lambda x)##. In the latter case, if ##\psi(x)## is a solution to the TISE, ##\psi(\lambda x)## won't be one, but you can still calculate expectation values.
Thank you. This point of Normalization seems so critical here. so I normalized ## \psi(\lambda x) ## and ## \sqrt {\lambda} ## was the Normalization constant and then I did the averaging again which yielded the correct answer which is ## \lambda^2<T> ##

## 1. What is a harmonic oscillator?

A harmonic oscillator is a type of system in which the restoring force is directly proportional to the displacement from the equilibrium position. This type of system exhibits repetitive motion and can be found in various physical systems such as pendulums, springs, and atoms.

## 2. How do kinetic energy and potential energy affect a harmonic oscillator?

In a harmonic oscillator, the kinetic energy (KE) and potential energy (PE) are constantly changing as the oscillator moves back and forth between its equilibrium position and its maximum displacement. KE is at its maximum when the oscillator is at the equilibrium position, while PE is at its maximum when the oscillator is at its maximum displacement.

## 3. What happens when the coordinates of a harmonic oscillator are re-scaled?

When the coordinates of a harmonic oscillator are re-scaled, it means that the system is being observed from a different reference frame or perspective. This does not change the physical properties of the oscillator, but it can affect the way we measure and analyze its KE and PE.

## 4. How does re-scaling of coordinates affect the KE and PE of a harmonic oscillator?

Re-scaling of coordinates does not affect the total amount of KE and PE in a harmonic oscillator, but it can change the ratio between the two. For example, if we observe the oscillator from a reference frame that is moving at a constant velocity, the KE will appear to be larger and the PE will appear to be smaller compared to observing it from a stationary reference frame.

## 5. What are some real-life examples of harmonic oscillators?

Some common examples of harmonic oscillators in real life include a swinging pendulum, a mass on a spring, and a vibrating guitar string. These systems exhibit repetitive motion and can be described by the same principles and equations as a simple harmonic oscillator in physics.

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