# How to deal with problems by changing the variables in wavefunctions?

I found a problems where a particle was moving under a potential $$V(r) = kr^2$$ and with a wave function $$\psi(r,t)$$. If the wave function is changed to $$\psi(ar ,t)$$, the ratio of average kinetic energy of final and initial is asked. Now if I take $$\psi$$ to be linear and take the $$a$$ out and then find kinetic energy the answer comes out to be wrong. In the solution they have taken $$ar$$ to be some dummy variable and then changed r to the dummy variable and then integrated it. Is taking $$\psi$$ linear wrong? And also what is the use of the potential given. Is it just to distract?

• Why would you expect $\psi$ to be linear in $r$? – J. Murray Mar 27 at 20:29
• It looks like they're trying to prove the virial theorem. In that case you should get the same answer if you take $\psi$ to be linear (after all, they prove it to be true for any function of $ar$ so it should work for the specific case of it being lienar). So in general your method should get the correct answer, but is just not a general result. – Alex Gower Mar 27 at 21:06

I think to answer your question, the point of the $$V(r) = kr^2$$ is not to distract, but because in one of the integrals you will need to make a change of variables $$R = ar$$ and you will get an $$a^{-2}$$ factor from the $$r^2$$.

For the broader picture, this technique is often used to show the virial theorem from the variational method in quantum mechanics (and alot of this method will be identical to what is wanted in your question).

The general proof of the virial theorem for a quadratic harmonic oscillator potential $$V(r) = kr^2$$ should go something like this:

Suppose an exact eigenstate of the system is $$\psi(\vec{r})$$ and consider a variational state $$\psi(a\vec{r})$$ (where we will set $$a=1$$ later but keep it general for now).

The expectation of the potential energy of the state in d-dimensions is given by:

$$\langle PE \rangle_{var} = \langle \psi|V(r)|\psi\rangle = \frac{\int d^d\vec{r} |\psi(a\vec{r})|^2 kr^2 }{\int d^d \vec{r} |\psi(a\vec{r})|^2}$$

(using the fact that $$\psi(a\vec{r})$$ will not be normalised so we must divide by its norm square in order to normalise the 2 $$\psi$$'s in the expectation value).

Using a change of variables $$\vec{R} = a\vec{r}$$ we get:

$$\langle PE \rangle_{var} = a^{-2} \frac{\int d^d\vec{R} |\psi(a\vec{R})|^2 kR^2 }{\int d^d \vec{r} |\psi(a\vec{R})|^2} = a^{-2} \langle PE \rangle_{exact}$$

(where we have seen that the fraction is equal to $$\langle PE \rangle_{exact}$$ since it's nothing but the usual expectation of potential energy for $$\psi(\vec{r})$$ but using the symbol $$\vec{R}$$ instead of $$\vec{r}$$)

Similarly we can show $$\langle KE \rangle_{var} = a^2 \langle KE \rangle_{exact}$$.

So $$\langle E \rangle_{var} = a^2 \langle KE \rangle_{exact} + a^{-2} \langle PE \rangle_{exact}$$

Therefore differentiating with respect to $$a$$ to find the optimum value of $$a$$ which minimises the total energy.

$$2a \langle KE \rangle_{exact} -2a^{-3} \langle PE \rangle_{exact} = 0$$

But we know $$a=1$$ is optimal so:

$$\langle KE \rangle_{exact} = \langle PE \rangle_{exact}$$

Try writing out the S.E.

$$$$\frac{-\hbar^2}{2m} \Delta\psi+kr^2\psi=E\psi$$$$

although cumbersome, it may be helpful to just solve the equation outright. In general it is not a good idea to assume the wavefunction is linear (this would usually make the wavefunction impossible to normalize over).

• Actually what they did was take the new wave function and operated on it using the H operator excluding the potential term which gives the final kinetic energy. I got the answer this way but I was confused because my first approach was to take it a linear function and take constant term out but I didn't understand why it was wrong. – aakash Mar 27 at 20:48