# Why must normalisable eigenfunctions have $E > V_{min}$? [duplicate]

I have read normalisable eigenfunctions of the Hamiltonian operator. $$\hat{H}\phi = E\phi$$ If $$\phi$$ is to be normalisable we must have $$E > V_{min}$$

Why is this?

• Do you know what the energies are for the Hydrogen atom wavefunctions? Are they normalisable? Commented Mar 4, 2021 at 16:08
• Where did you read this? It's not true. Commented Mar 4, 2021 at 16:19
• @Javier I have corrected the question, within the context that I read it originally $V_{min} = 0$ so they just said $E > 0$ but they meant $E > V_{min}$ Commented Mar 4, 2021 at 16:22
• Does this answer your question? Lower bound on energy is potential minimum Commented Mar 4, 2021 at 16:28

It really is 'just mathematics'.

The solutions of the Time Independent Schrödinger Equation $$\hat{H}\psi = E\psi$$ are such that the lowest allowed energy level ($$E_1$$) is always non-zero.

It's a property that distinguishes quantum systems sharply from Classical systems.

Please note that the OP altered his question substantially since this answer was formulated.

• But you could always add a constant to the Hamiltonian which could result in a zero ground state energy, no? Commented Mar 4, 2021 at 16:25
• As far as I know, the Hamiltonian is required to be bounded from below (such that there exists a finite ground state energy). Adding a constant won't change the physics but shifts the spectrum such that E_1=0 (for example). Commented Mar 4, 2021 at 16:29
• Afaik you can shift any observable by a constant without affecting it's legitimacy. The identity is Hermitian so you're not going to lose the most important property. It is common for instance in deriving the uncertainty relation for two operators to shift them by a constant so that their mean is zero. Commented Mar 4, 2021 at 16:37
• @Gert (this may not be the most rigorous proof, but anyway): Take any Hamiltonian which satisfies $\hat{H} |\phi_i\rangle = E_i\, |\phi_i\rangle$ with $E_i$ ordered increasingly. The Hamiltonian $\hat{h} \equiv \hat{H}-E_1$ fulfillls $\hat{h}|\phi_i\rangle = (E_i-E_1)|\phi_i\rangle$. Note that $(E_i-E_1) \geq 0$ for all $i$. Commented Mar 4, 2021 at 16:41
• @Jakob This is becoming 'how many angels on a pin'? Have the last word by all means. Energy is relative to some reference point, no dispute here. Have a nice day now.
– Gert
Commented Mar 4, 2021 at 16:48