Lower bound on energy is potential minimum

Suppose we have a particle of mass $m$ that is in an eigenstate $|\psi\rangle$ of the Hamiltonian $\hat{H}=\hat{T}+\hat{V}$, where $\hat{T}$ is the kinetic energy operator and $\hat{V}=V(\hat{r})$ is the potential energy operator. If the potential has lower bound $V_0$, then is it necessary for the energy eigenvalue $E$ of $|\psi\rangle$ to be greater than $V_0$? Classically this is true, since we regard negative kinetic energy as physically unrealizable/meaningless. However, I don't know if I can necessarily say the same in the quantum case. For example, what I'm tempted to do is write

$$\langle\psi | \hat{T}|\psi \rangle = \langle\psi | E-\hat{V}|\psi\rangle$$

and say "If $E<V_0$, then the RHS is necessarily negative, implying the LHS is as well, which we will regard as physically meaningless. If $E=V_0$, then $|\psi\rangle$ is trivial.", but I'm not sure if that's right.

Confused by that, I then wanted to show that, if $E< V_0$, then a non-trivial $|\psi\rangle$ is non-normalizable. However, I'm not entirely sure how to do this.

• @AccidentalFourierTransform Oh sorry no, after looking a bit more into it I think your answer is fine. I just always wait a while before accepting an answer. :) Commented Mar 27, 2016 at 18:20

The operator $\hat T$ is positive definite (a.e), which means that for most kets $|\varphi\rangle\neq 0$ you have $$\langle \varphi|\hat T|\varphi\rangle>0$$

One way to see this is that $\hat T$ is quadratic in $\hat P$, which is itself self-adjoint. Therefore $$\langle\varphi|\hat T|\varphi\rangle=\frac{1}{2m}\langle\varphi|\hat P^2|\varphi\rangle=\frac{1}{2m}\big|\big|\hat P|\varphi\rangle\big|\big|^2>0$$

Alternatively, we know that $\hat T$ is proportional to the Laplacian $-\Delta$, which is positive-definite (a.e.), for example see The minus Laplacian operator is positive definite.

With this, it is easy to see that $$E=\langle\varphi|\hat T+\hat V|\varphi\rangle\ge\langle\varphi|\hat V|\varphi\rangle\ge V_0\langle\varphi|\varphi\rangle=V_0$$

• Sorry for being late but why does $||\hat{P}|\varphi\rangle||>0$ hold a.e.? Commented Jun 26, 2021 at 14:41
• @test123 Because $||\cdot||$ is a norm, it is positive definite. It is strictly positive unless the argument is the zero vector. Commented Jun 27, 2021 at 19:16
• Thank you for the reply but this would imply $\hat{P}|\psi\rangle\neq 0$ a.e. so $\hat{P}$ is a.e. injective. Is this perhaps some theorem from functional analysis? Commented Jun 28, 2021 at 8:05
• @test123 By $\hat P|\psi\rangle\neq0$ a.e. I mean that, for most kets $|\psi\rangle$, the ket $\hat P|\psi\rangle$ is non-zero. This is not any deep theorem, just the statement that most functions are not constant. Commented Jun 28, 2021 at 17:40
• @test123 I think you are expecting a deeper point than the one I am trying to make. Pick a random function. What is the probability that it is the constant function? The answer is zero. Most functions are not even continuous, let alone constant! Commented Jul 7, 2021 at 10:55