- #1
JTFreitas
- 18
- 3
- TL;DR Summary
- I am trying to find an intuitive way of obtaining the zero-point energy of the harmonic oscillator based on the Hamiltonian and my knowledge of linear algebra.
First time posting in this part of the website, I apologize in advance if my formatting is off.
This isn't quite a homework question so much as me trying to reason through the work in a way that quickly makes sense in my head. I am posting in hopes that someone can tell me if my reasoning is sound or not.
Say we have a harmonic oscillator, and we establish the usual ladder operators, ##\hat{a}, \hat{a}^{\dagger}##
Then we can write the Hamiltonian of the system as
$$ \hat{H} = \hbar\omega\left(\hat{a}^{\dagger}\hat{a} + \frac{1}{2}\right)$$
And it follows right away that the Hamiltonian of some state:
$$ \hat{H}\ket{n} = \hbar\omega\left(\hat{a}^{\dagger}\hat{a} + \frac{1}{2}\right)\ket{n}$$
Let's say this is all I know at this point, I did no other work of trying to solve the Schrödinger equation, any sort of analysis, nothing else.
(1) However, I do know from linear algebra that there is a little something called the kernel. So I suppose that we're considering a particular ##\ket{n}## that makes up ##Ker(\hat{a})##, that is ##\hat{a}\ket{n} = \mathbf{0}##
Then, the eigenvalue problem reduces to
$$ \hat{H}\ket{n} = \frac{\hbar\omega}{2}\ket{n} $$
(2) And through my knowledge that the expected value of the energy of the system must be larger than 0, it follows directly that the minimum energy of the system is ##E_o = \frac{1}{2}\hbar\omega##, since this is the smallest eigenvalue that the Hamiltonian accepts.
What I'm unsure about: Is it safe to assume that ##\hat{a}## has a non-empty kernel? Is it just something we establish, or do I need to justify (1) with better arguments? If so, how could I do that? Is (2) anyhow related with my assumption of (1), does one follow from the other?
This isn't quite a homework question so much as me trying to reason through the work in a way that quickly makes sense in my head. I am posting in hopes that someone can tell me if my reasoning is sound or not.
Say we have a harmonic oscillator, and we establish the usual ladder operators, ##\hat{a}, \hat{a}^{\dagger}##
Then we can write the Hamiltonian of the system as
$$ \hat{H} = \hbar\omega\left(\hat{a}^{\dagger}\hat{a} + \frac{1}{2}\right)$$
And it follows right away that the Hamiltonian of some state:
$$ \hat{H}\ket{n} = \hbar\omega\left(\hat{a}^{\dagger}\hat{a} + \frac{1}{2}\right)\ket{n}$$
Let's say this is all I know at this point, I did no other work of trying to solve the Schrödinger equation, any sort of analysis, nothing else.
(1) However, I do know from linear algebra that there is a little something called the kernel. So I suppose that we're considering a particular ##\ket{n}## that makes up ##Ker(\hat{a})##, that is ##\hat{a}\ket{n} = \mathbf{0}##
Then, the eigenvalue problem reduces to
$$ \hat{H}\ket{n} = \frac{\hbar\omega}{2}\ket{n} $$
(2) And through my knowledge that the expected value of the energy of the system must be larger than 0, it follows directly that the minimum energy of the system is ##E_o = \frac{1}{2}\hbar\omega##, since this is the smallest eigenvalue that the Hamiltonian accepts.
What I'm unsure about: Is it safe to assume that ##\hat{a}## has a non-empty kernel? Is it just something we establish, or do I need to justify (1) with better arguments? If so, how could I do that? Is (2) anyhow related with my assumption of (1), does one follow from the other?