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I just stumbled across the problem and have no idea how to solve it: "Considering the Time-Independent Schrodinger Equation for a stationary state $\psi$ with energy $E$, that is $$\psi '' = \frac{2m}{\hbar ^2 }(V(x)-E)\psi,$$ prove that $E$ must exceed the minimum value of the potential, by noting that $E=\left\langle H \right\rangle $"

My attempt:

We assume firstly that $E<\min(V(x))$, and thus $V(x)-E>0$ for all $x$. Thus if define $\frac{2m}{\hbar ^2 }(V(x)-E) = A(x)>0$, as all the terms in the function are positive, we get: $$\psi '' = A(x)\psi$$ Since $A(x)>0$, we know that $\psi ''$ and $\psi$ have the same sign. I don't know where to go from here. Haven't even used the fact that $E=\left\langle H \right\rangle $. Any help would be appreciated!

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    $\begingroup$ Just write down the formula for $\langle H\rangle$ as the problem hints. Can $\langle H\rangle$ possibly be below $V_{\rm min}$? $\endgroup$
    – mike stone
    Feb 27, 2021 at 22:40
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    $\begingroup$ Your question was answered here already. $\endgroup$ Feb 27, 2021 at 22:51
  • $\begingroup$ @mikestone I tried to do that but I only get a trivial result. H acting on the wavefunction yields an E and assuming the state is normalized I get that <H>=E. Haven't been able to get anything out using the hermiticity of H either. $\endgroup$
    – Johnn.27
    Feb 27, 2021 at 22:58
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    $\begingroup$ @NicolásMaílloGómez When following the hint to write down $\langle H \rangle$, don't use $H | \psi \rangle= E |\psi\rangle$ (which is only valid for an eigenket). Use $H=p^2/(2m)+V$, so $\langle H \rangle = \langle p^2/(2m) \rangle + \langle V \rangle$. Then try to think about what inequalities you can use. Can both terms here be positive or negative, or is one non-negative only? Are there any other lower bounds / minimum values you can think of that might be useful in proving $\langle H \rangle \geq V_{\rm min}$? $\endgroup$
    – Andrew
    Feb 28, 2021 at 0:58

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$\require{cancel}$I am pretty sure you can demonstrate this with contradiction. You can however prove this directly as follows: Let $\psi(x)$ be a real and normalized eigenfunction of the Hamiltonian $H$. It is a known theorem that the eigenvalue $E$ that corresponds to $\psi(x)$ is the expected value of $H$ on the eigenfunction itself. Thus, $$E = \langle H\rangle \Longleftrightarrow E = \int_{\mathbb{R}} \psi(x) \left( -\frac{\hbar^2}{2m} \frac{\text{d}^2}{\text{d}x^2} + V(x) \right) \psi(x) \, dx \Longrightarrow $$ $$E = - \frac{\hbar^2}{2m} \int_{\mathbb{R}} \psi(x) \psi''(x) \, dx + \int_{\mathbb{R}} V(x)\psi^2(x) \, dx$$ By using integration by parts on the first integral, we obtain: $$E = -\frac{\hbar^2}{2m} \left( \bcancel{\left[ \psi'(x) \psi(x) \right]_{-\infty}^{+\infty}} - \int_{\mathbb{R}} \left(\frac{\text{d} \psi}{\text{d}x} \right)^2 \, dx \right) + \int_{\mathbb{R}} V(x) \psi^2(x) \, dx \Longrightarrow $$ $$E = + \frac{\hbar^2}{2m} \int_{\mathbb{R}} \left(\frac{\text{d} \psi}{\text{d}x} \right)^2 \, dx + \int_{\mathbb{R}} V(x) \psi^2(x) \, dx$$ Since $\psi''(x)^2 \geqslant 0$, we have: $$\frac{\hbar^2}{2m} \int_{\mathbb{R}} \left(\frac{\text{d} \psi}{\text{d}x} \right)^2 \, dx \geqslant 0$$ and thus, $$E \geqslant \int_{\mathbb{R}} V(x) \psi^2(x) \, dx$$ Finally, since $V(x)$ exhibits a minimum $\min V$, we have: $$V(x) \geqslant \min V \Longrightarrow V(x) \psi^2(x) \geqslant (\min V) \cdot \psi^2(x) \Longrightarrow $$ $$ \int_{\mathbb{R}} V(x) \psi^2(x) \, dx \geqslant \int_{\mathbb{R}} (\min V) \cdot \psi^2(x) = \min V \text{ since } \int_{\mathbb{R}} \psi^2(x) \, dx = 1$$ Consequently, $$E \geqslant \int_{\mathbb{R}} V(x) \psi^2(x) \, dx \geqslant \min V \Longrightarrow E \geqslant \min V $$ Now, there is an intuitive reason why this happens (that is, the eigenvalues are bounded below from the minimum value of the potential function). You very well demonstrated that the converse of the statement would imply $\psi(x)$ and $\psi''(x)$ to have the same sign in $\mathbb{R}$ (or perhaps zero, but it's not of interest here). That means, that if $\psi(x)$ is positive, so is $\psi''(x)$. The second derivative of $\psi(x)$ however dictates the concavity of the $\psi(x)$. This condition renders $\psi(x)$ incapable of being normalizable in $\mathbb{R}$. You can perhaps see this by drawing a few graphs yourself. This is the argument Griffiths persuades the reader to make in his book Introduction to Quantum Mechanics (The last exercise at the end of Chapter 2).

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    $\begingroup$ Actually, the direct proof is the only general proof. The intuitive reason fails for wavefunction on a compact support. $\endgroup$ Feb 28, 2021 at 5:55
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The equation $\psi'' = +A(x)\psi$ can be interpreted as the curvature of the wavefunction, $\psi''$, having the same sign as the wavefunction, $\psi$.

That means when the wavefunction is positive, $\psi > 0$, the function has positive curvature, $\psi'' > 0$, like a smile. But that implies as we go off towards positive and negative infinity in the $x$ direction, the wavefunction is growing away from the $x$ axis.

If the wavefunction is negative, $\psi < 0$, then the curvature is also negative, $\psi''<0$, like a frown. And the wavefunction grows towards negative infinity as we go out on the $x$ axis.

In either case, the wavefunction is not normalizable. We conclude if $E<V_{min}$, the wavefunction cannot be normalized. This consequently makes expectation values like $\langle H \rangle$ undefined by their very definition,

$$ \langle H \rangle = \langle \psi | \hat{H} | \psi \rangle = \int_{-\infty}^{+\infty} \psi^*(x) \ \bigg{(} \hat{H} \ \psi(x) \bigg{)} \ dx = \int_{-\infty}^{+\infty} \psi^*(x) \ E_ \ \psi(x) \ dx$$

$$ = E\int_{-\infty}^{+\infty} \psi^*(x) \ \psi(x) \ dx \neq \text{finite}$$

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  • $\begingroup$ This answer is useless if the support of $\psi$ is not infinite like the case of the infinite potential well or the angular dependence in more than 1D. $\endgroup$ Feb 28, 2021 at 5:51

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