6
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • $\begingroup$ @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. :) $\endgroup$
    – Arturo don Juan
    Commented Mar 27, 2016 at 18:20

1 Answer 1

Reset to default
11
$\begingroup$

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 $$ E=\langle\varphi|\hat T+\hat V|\varphi\rangle\ge\langle\varphi|\hat V|\varphi\rangle\ge V_0\langle\varphi|\varphi\rangle=V_0 $$

Improve this answer
$\endgroup$
6
  • $\begingroup$ Sorry for being late but why does $||\hat{P}|\varphi\rangle||>0$ hold a.e.? $\endgroup$
    – test123
    Commented Jun 26, 2021 at 14:41
  • $\begingroup$ @test123 Because $||\cdot||$ is a norm, it is positive definite. It is strictly positive unless the argument is the zero vector. $\endgroup$
    – AccidentalFourierTransform
    Commented Jun 27, 2021 at 19:16
  • $\begingroup$ 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? $\endgroup$
    – test123
    Commented Jun 28, 2021 at 8:05
  • $\begingroup$ @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. $\endgroup$
    – AccidentalFourierTransform
    Commented Jun 28, 2021 at 17:40
  • 1
    $\begingroup$ @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! $\endgroup$
    – AccidentalFourierTransform
    Commented Jul 7, 2021 at 10:55

Not the answer you're looking for? Browse other questions tagged or ask your own question.