# Ground state 1d problems

1. Mar 16, 2013

### LagrangeEuler

Is there some mathematical prove that for ground state in QM problems wave function doesn't have any zeros?

2. Mar 16, 2013

Staff Emeritus
Yes there is.

3. Mar 16, 2013

### LagrangeEuler

Could you tell me some reference or link where I can find it.

4. Mar 16, 2013

Staff Emeritus
You can prove it yourself. Look at the time-independent energy operator, and ask what that tells you about the slope of the wavefunction. Then, ask yourself what nodes in the wavefunction mean to the slopes.

5. Mar 16, 2013

### LagrangeEuler

I have shown that for infinite square well and linear harmonic oscillator, but I don't know how to show that for any potential $V(x)$.
$$\frac{d^2\phi}{dx^2}+\frac{2m}{\hbar^2}(E-V(x))\phi(x)=0$$
That means that $\frac{d\phi}{dx}=0$, $\frac{d^2\phi}{dx^2}<0$

6. Mar 16, 2013

Staff Emeritus
Hint: deivative of the wavefunction.

7. Mar 16, 2013

### LagrangeEuler

I don't understand. Wave function in ground state has one maximum. So for some $x=a$, $\frac{d\phi}{dx}|_{x=a}=0$, $\frac{d^2\phi}{dx^2}|_{x=a}<0$ and point $a$ is unique in the whole region. Right? I think that is the logic, but I still don't know how to prove that.

Last edited: Mar 16, 2013
8. Mar 16, 2013

Staff Emeritus
Hint: average value of the derivative of the wavefunction

9. Mar 16, 2013

### LagrangeEuler

Average value of derivative of wave function is zero. Right?
$$\frac{1}{b-a}\int^b_a \frac{d\psi}{dx}dx=\frac{1}{b-a}(\psi(b)-\psi(a))$$
This is result for any state, and not just for ground state.

10. Mar 16, 2013

Staff Emeritus
OK, absolute value...or square. I'm trying to get you to think, not calculate.

11. Mar 16, 2013

### LagrangeEuler

I'm trying. Ground state must have one stationary point $\frac{d\phi}{dx}|_{x=a}=0$. All other states has more then one stationary point. That stationary point is maximum, so $\frac{d^2\phi}{dx^2}|_{x=a}<0$. From some point to $a$ function $\phi(x)$ increase, and then decrease. Maybe u want me to see that the point $a$ is in the middle of this two intervals. That's the case in the problem that I saw till now.

12. Mar 21, 2013

### LagrangeEuler

I tried to prove but I still don't know how. Tnx for your answer.

13. Mar 21, 2013

### stevendaryl

Staff Emeritus
14. Mar 22, 2013

### Avodyne

Suppose you have a wave function with a simple zero at x=a. Thus, near x=a we can write ψ(x)=c*(x-a), where c is a constant.

Let ε be a small length, and consider the following alternative wave function:
χ(x) = -ψ(x) for x < a-ε
χ(x) = cε (that is, constant) for a-ε < x < a+ε
χ(x) = +ψ(x) for x > a+ε

Compute the expectation value of the hamiltonian in ψ and in χ, and show that the expectation value in χ is smaller than it is in ψ. This implies that ψ cannot be the ground state, because the ground state minimizes the expectation value of H. Therefore, ψ cannot have a zero.

15. Mar 22, 2013

### stevendaryl

Staff Emeritus
Very nice argument.

16. Mar 22, 2013

### DrDu

I think the book by Messiah, Quantum Mechanics, also has a nice proof based on the Wronskian.