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

[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?

 

[kuruman]

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 $\lambda =1$.

 

[Baibhab Bose]

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?

 

[kuruman]

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.

 

[Baibhab Bose]

Oh, that's a bad mistake!
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..?

 

[kuruman]

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.

 

[Baibhab Bose]

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
$\psi(u)$ ?

 

[kuruman]

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.

 

[Baibhab Bose]

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?

 

[kuruman]

When I wrote

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.

 

[vela]

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.

 

[Baibhab Bose]

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 $\lambda^2<T>$