Does the ground state wave function in QM problems have any zeros?

[LagrangeEuler]

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

 

[Vanadium 50]

Yes there is.

 

[LagrangeEuler]

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

 

[Vanadium 50]

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.

 

[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 $\frac{d\phi}{dx}=0$, $\frac{d^2\phi}{dx^2}<0$

 

[Vanadium 50]

Hint: deivative of the wavefunction.

 

[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.

 

[Vanadium 50]

Hint: average value of the derivative of the wavefunction

 

[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.

 

[Vanadium 50]

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

 

[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.

 

[LagrangeEuler]

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

 

[stevendaryl]

There is an argument given in this paper:View attachment nodes.pdf

 

[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.

 

[stevendaryl]

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.

Very nice argument.

 

[DrDu]

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