Proof of nodes of bound states using the wronskian

Question

I've been trying to solve this problem for a couple of weeks now, but I don't seem to get nowhere with it. I tried to prove it by contradiction, supposing $\psi_q$ has no nodes, and finding some equality involving $\psi_q$ where the other side of the equation will have to change sign, but I can't find it.

I also tried to find some differential equation for the Wronskian, and from then derive an explicit formula, but I go nowhere. I also tried to use the limits and the normalization of the wavefunctions (the fact that they will approach zero as $x$ approaches infinity), but it gets me nowhere. Can someone help with this?

The Wronskian of two functions $\psi_p$ and $\psi_q$ is defined as: $$ W_{p,q} = \psi_p \psi_q' - \psi_q \psi_p'. $$

Now consider the following question:

We now consider two bound states $\psi_p$ and $\psi_q$, with $q > p $. Using the Wronskian, show that:

• a) $\psi_q$ has at least one zero;

• b) If $\psi_p$ has a zero, the smallest zero of $\psi_q$ is strictly less than the smallest zero of $\psi_p$, and the largest zero of $\psi_q$ is strictly greater than the largest zero of $\psi_p$.

Additionally, we have the following relation involving the derivative of the Wronskian: $$ W'_{p,q} = \frac{2m}{\hbar^2} (E_q - E_p) \psi_p \psi_q. $$ edit: I tried something, but I'm still having doubts. Here's my reasoning.

We want to compute the following integral: $$ \ \int_{-\infty}^{+\infty} \frac{d}{dx} W(\psi_p, \psi_q)(x) \, dx = \frac{2m}{\hbar^2} \left( E_q - E_p \right) \int_{-\infty}^{+\infty} \psi_p(x) \psi_q(x) \, dx \ $$

$$ \int_{-\infty}^{+\infty} \left( \frac{\hbar^2}{2m} (E_q - E_p) \psi_p(x) \psi_q(x) \right) \, dx = 0 $$

By using the fact that the functions $\psi_p(x)$ and $\psi_q(x)$ are orthogonal (i.e., their integral vanishes), we are left with:

$$ \int_{-\infty}^{+\infty} \frac{d}{dx} W(\psi_p, \psi_q)(x) \, dx = 0 $$

Thus, the Wronskian $W(\psi_p, \psi_q)(x)$ must be constant, and its limit as $ x \to \pm \infty$ must be zero, implying:

$$ W(\psi_p, \psi_q)(x) = 0 \quad \text{for all} \, x $$

First, I want to make sure of something. The fact that the wronskian is zero doesn't imply that the wavefucntions are linearly dependant. actually, if the wavefucntions are linearly dependant,this implies the wronskian equal zero, right ?

Second, I feel like I wasn't rigorous when I calculated the integral of the derivative of the wronskian. can the result be that the wronskian is constant, or actually it's just proves that the limits in inf are equal.

Can someone check if the reasoning is rigorous?

Answer 1

First, I want to make sure of something. The fact that the wronskian is zero doesn't imply that the wavefucntions are linearly dependant. actually, if the wavefucntions are linearly dependant,this implies the wronskian equal zero, right ?

That is not so simple. You could check in the Wronskian wiki some criteria that, considered together with the nullity of the Wronskian, give us linear dependence. One of them is if the functions are analytic over the whole interval where the Wronskian vanishes, which is your case.

But this is good for you!

Second, I feel like I wasn't rigorous when I calculated the integral of the derivative of the wronskian.

In fact, there is one missing argument. The property

$$ \int_{-\infty}^{\infty} f(x)dx = 0 \Rightarrow f(x)=0, \,\, \forall x $$

Is true only if $f(x)$ never change it's signal in its domain. In your case, it could be true if you assume something...

I think you're in the right way. I decide to post it as an answer, otherwise it would be a lot of comments. But I think it would be nice to see you answering your own question at the final. All the best!.

Answer 2

Start with $$ \hat H=-\frac{\hbar ^{2}}{2m}\frac{d^{2}}{d\xi ^{2}}+V(\xi ). $$

The node theorem states that, if $$ \hat H\psi _{1}(\xi )=E_{1}\psi _{1}(\xi )\, ,\qquad \hat H\psi_{2}(\xi)=E_{2}\psi _{2}(\xi)\, ,\qquad E_{1}<E_{2}, $$ then $\psi _{2}(\xi )$ has more zeroes (nodes) than $\psi _{1}(\xi ).$

To show this, manipulate the Schrödinger equation so that $$ \frac{1}{\psi _{1}(\xi )}\frac{\hbar ^{2}}{2m}\frac{d^{2}}{d\xi ^{2}}\psi _{1}(\xi )+E_{1}=V(\xi )=\frac{1}{\psi _{2}(\xi )}\frac{\hbar ^{2}}{2m}\frac{% d^{2}}{d\xi ^{2}}\psi _{2}(\xi )+E_{2}, $$ which one then reorganizes as \begin{align} \psi _{2}(\xi )\frac{d^{2}}{d\xi ^{2}}\psi _{1}(\xi )-\psi _{1}(\xi )\frac{d^{2}}{d\xi ^{2}}\psi _{2}(\xi ) &=\frac{d}{d\xi }\left( \psi _{2}(\xi ) \frac{d}{d\xi }\psi _{1}(\xi )-\psi _{1}(\xi )\frac{d}{d\xi }\psi _{2}(\xi )\right) \\ &=\frac{\hbar ^{2}}{2m}\left( E_{2}-E_{1}\right) \psi _{1}(\xi )\psi _{2}(\xi ). \end{align} Integrating from $\xi =a$ to $\xi =b,$ we find \begin{align} &\frac{\hbar ^{2}}{2m}\left( E_{2}-E_{1}\right) \int_{a}^{b}d\xi \psi _{1}(\xi )\psi _{2}(\xi ) \\ \qquad &= \left( \psi _{2}(b)\frac{d}{d\xi }\psi _{1}(b)-\psi _{1}(b)\frac{d}{d\xi }\psi _{2}(b)\right) \\ &\qquad\qquad -\left( \psi _{2}(a)\frac{d}{d\xi }\psi _{1}(a)-\psi _{1}(a)\frac{d}{d\xi } \psi _{2}(a)\right) \tag{1} \end{align} If one chooses the points $a,$ $b$ to be consecutive zeroes of $\psi _{1},$ and if $\psi _{1}$ is continuous, then $\psi _{1}$does not change sign over the interval $\left[ a,b\right] .$ We can assume, without loss of generality, that $\psi _{1}$ is positive on this interval, so that $\psi _{1}^{\prime }(a)\geq 0$ and $\psi _{1}^{\prime }(b)\leq 0.$ Then, using these with $\psi _{1}(a)=\psi _{1}(b)=0$ in Eqn.(1), we obtain $$ \frac{\hbar ^{2}}{2m}\left( E_{2}-E_{1}\right) \int_{a}^{b}d\xi \psi _{1}(\xi )\psi _{2}(\xi )=\psi _{2}(b)\frac{d}{d\xi }\psi _{1}(b)-\psi _{2}(a)\frac{d}{d\xi }\psi _{1}(a). $$

Assume now that $\psi _{2}(\xi )$ is always positive on $\left[ a,b\right].$ Then the integral on the left is positive, so the integral will be positive (the area under a positive curve is necessarily positive!) On the right hand side, however, the first term is negative by virtue of $\psi _{1}^{\prime }(b)\leq 0$ while the second is positive. Overall, the right hand side is therefore negative, while the left side is positive, a contradiction : $\psi _{2}(\xi ) $ cannot be always positive on $\left[ a,b\right]$, meaning that between two consecutive zeroes of $\psi _{1}(\xi )$, the function $\psi _{2}(\xi )$ must itself go through zero. In other words, $\psi _{2}(\xi )$ has more zeroes than $\psi _{1}(\xi ).$ This proves the theorem.