I Zero-point energy of the harmonic oscillator

[JTFreitas]

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?

 

[PeroK]

In general an operator may have a zero kernel. You have to prove, therefore, that the lowering operator has a non-zero kernel.

PS The raising operator must have zero kernel, for example.

 

[JTFreitas]

In general an operator may have a zero kernel. You have to prove, therefore, that the lowering operator has a non-zero kernel.

PS The raising operator must have zero kernel, for example.

Okay, that makes sense. Thanks for your answer.

I suppose there is no direct mathematical way of establishing the kernels of these particular operators then? Other than writing them out in their matrix representation, finding the inverse, solving $\hat{a}\mathbf{x} = 0$, etc etc etc...?

 

[vanhees71]

The eigenstates of the annihilation operator are the coherent states (Glauber states in quantum optics):

https://en.wikipedia.org/wiki/Coherent_state

See also the more general related case of squeezed states:

https://en.wikipedia.org/wiki/Squeezed_coherent_state

 

[PeroK]

Okay, that makes sense. Thanks for your answer.

I suppose there is no direct mathematical way of establishing the kernels of these particular operators then? Other than writing them out in their matrix representation, finding the inverse, solving $\hat{a}\mathbf{x} = 0$, etc etc etc...?

You can prove that the raising operator has zero kernel by:
$$\hat a^{\dagger}|\alpha \rangle = 0 \ \Rightarrow \ \hat H|\alpha \rangle = \hbar\omega\left(\hat a \hat{a}^{\dagger} - \frac{1}{2}\right) |\alpha \rangle = -\frac{\hbar\omega}{2}|\alpha \rangle$$ And, as you cannot have negative energy eigenstates, it follows that $|\alpha \rangle = 0$.

In fact, the proof that the eigenvalues have these half-integer values follows similarly. If we start with any eigenstate and repeatedly apply the lowering operator, then we get a sequence of eigenstates with lower and lower energies. Eventually, unless we hit the zero state, we get negative energies. This implies that there must be some state(the ground state $|0 \rangle$) that is in the kernel of $\hat a$: $$\hat a|0\rangle = 0$$

 

[JTFreitas]

You can prove that the raising operator has zero kernel by:
$$\hat a^{\dagger}|\alpha \rangle = 0 \ \Rightarrow \ \hat H|\alpha \rangle = \hbar\omega\left(\hat a \hat{a}^{\dagger} - \frac{1}{2}\right) |\alpha \rangle = -\frac{\hbar\omega}{2}|\alpha \rangle$$ And, as you cannot have negative energy eigenstates, it follows that $|\alpha \rangle = 0$.

In fact, the proof that the eigenvalues have these half-integer values follows similarly. If we start with any eigenstate and repeatedly apply the lowering operator, then we get a sequence of eigenstates with lower and lower energies. Eventually, unless we hit the zero state, we get negative energies. This implies that there must be some state(the ground state $|0 \rangle$) that is in the kernel of $\hat a^{\dagger}$: $$\hat a^{\dagger}|0\rangle = 0$$

Once again, thanks for the answer, this is really helpful.

Then, this does mean that both kernels can be deduced based on the very physical properties of the system. As in, the zero-kernel of the raising operator, and the non-zero kernel of the lowering operator are necessary conditions for the hamiltonian to represent a "real" physical system (quotes on "real" because I'm unsure if I'm not misusing the word.)

The eigenstates of the annihilation operator are the coherent states (Glauber states in quantum optics):

https://en.wikipedia.org/wiki/Coherent_state

See also the more general related case of squeezed states:

https://en.wikipedia.org/wiki/Squeezed_coherent_state

These resources have also been a really good addition to the picture I'm trying to paint in my head!

Thank you both for the really helpful responses!

 

[PeroK]

Then, this does mean that both kernels can be deduced based on the very physical properties of the system. As in, the zero-kernel of the raising operator, and the non-zero kernel of the lowering operator are necessary conditions for the hamiltonian to represent a "real" physical system (quotes on "real" because I'm unsure if I'm not misusing the word.)

Yes. There are two extraordinary examples in QM of the power of linear algebra/functional analysis: the algebraic development of the Quantum Harmonic Oscillator; and, the algebraic theory of Quantum Angular Momentum. In both cases a minimal set of physical assumptions leads to a full algebraic development of the theory. You might say that the same is true of the QM mathematical formalism in general.

 

[PeterDonis]

Eventually, unless we hit the zero state, we get negative energies. This implies that there must be some state(the ground state $|0 \rangle$) that is in the kernel of $\hat a^{\dagger}$: $$\hat a^{\dagger}|0\rangle = 0$$

I think you mean $a$ here (the annihilation/lowering operator), not $a^\dagger$ (the creation/ raising operator).

 

[PeroK]

I think you mean $a$ here (the annihilation/lowering operator), not $a^\dagger$ (the creation/ raising operator).

Yes, thanks, I forgot to take the daggers off.

 

[vanhees71]

Yes. There are two extraordinary examples in QM of the power of linear algebra/functional analysis: the algebraic development of the Quantum Harmonic Oscillator; and, the algebraic theory of Quantum Angular Momentum. In both cases a minimal set of physical assumptions leads to a full algebraic development of the theory. You might say that the same is true of the QM mathematical formalism in general.

The same is true for the hydrogen atom (the usual non-relativistic theory without fine structure). Thanks to the dynamical symmetry with the Runge-Lenz vector as an additional conserved quantity you can get the energy eigenvalue problem (i.e., the usual simultaneous diagonalization of $\hat{H}$, $\hat{\vec{L}}^2$, and $\hat{L}_3$) by using the representation theory for the groups O(4) (for $E<0$), which can be reduced to the usual angular-momentum representations, ISO(3) (for $E=0$), which is known from the free particle, and $\text{SO}(1,3)^{\text{\uparrow}}$, which can also reduced to the angular-momentum representations. That's in fact the way, how Pauli solved the hydrogen problem for the first time within the matrix-mechanics formulation by Born, Jordan, and Heisenberg (before Schrödinger solved it within his wave-mechanics formalism, which was a great relief for most of the physicists at the time, because the algebraic methods were much less known than the formalism to solve partial differential equations).