# SHO Ladder Method missed states?

1. Jan 17, 2012

### aimforclarity

Following Griffiths derivation on pg 44 of the eigen-states of SHO Hamiltonian, he says that we can now find all eigenvalues, but doesnt say how he knows that a and a dagger will indeed take you between nearest neighboring orthogonal states.

in other words, how do we know the ladder operators does raise us by quanta of energy,

or equivalently, how do we know that hbar omega is the smallest quanta of energy give the SHO Hamiltonian ?

Thank you :)

Aim For Clarity

2. Jan 18, 2012

### Bill_K

1) H = a*a + ½. This means that any energy eigenstate is an eigenstate of the number operator N = a*a.
2) Given any energy eigenstate |E> with eigenvalue E, show that an|E> is also an eigenstate with eigenvalue E - n. Thus we can construct a sequence of eigenstates with progressively lower energy.
3) There will fail to be a ground state unless the sequence terminates. From the normalization, one finds that the sequence will terminate only if E - n = 0 for some n. Therefore the only possible eigenvalues are zero and the positive integers, and (a*)n|0> are the only eigenstates.

3. Jan 21, 2012

### aimforclarity

this is all true, and is a summary of grifits derivation, but it it doesn't tell you that the raising and lowering operators take you between nearest orthogonal states. what is they raised and lowered by two quanta of energy,

how do you know what the smallest quanta of energy is?

4. Jan 21, 2012

### jewbinson

Given (2) to be true, if we have another eigenstate of H with eigenvalue NOT equal to hw(n+1/2) where n is an integer, then we get a problem. Can you see what this problem is?

(so for example, say we have an eigenstate of H with eigenvalue (4 + 1/3)hw. Try and get a contradiction.)

5. Jan 21, 2012

### Dickfore

It follows from the fact that the eignevalues of the hermitian operator $\hat{n} = \hat{a}^{\dagger} \, \hat{a}$ are non-negative! Namely, let:
$$\hat{n} \, \vert \nu \rangle = \nu \, \vert \nu \rangle$$
Assuming the eigenket $\vert \nu \rangle$ is normalized to unity ($\langle \nu \vert \nu \rangle = 1$), we have:
$$\nu = \langle \nu \vert \hat{n} \vert \nu \rangle = \langle \nu \vert \hat{a}^{\dagger} \, \hat{a} \vert \nu \rangle$$
Now, insert a compete orthonormal basis $\lbrace k' \rbrace$ between $\hat{a}$ and $\hat{a}^\dagger$:
$$\nu = \sum_{k'}{\langle \nu \vert \hat{a}^{\dagger} \vert k' \rangle \, \langle k' \vert \hat{a} \vert \nu \rangle} = \sum_{k'}{\left\vert \langle k' \vert \hat{a} \vert \nu \rangle \right\vert^2}$$
Each of the summands is a non-zero real number, being a modulus of a complex number. Therefore, $\nu \ge 0$.

Now, from the commutation relation $\left[ \hat{a}, \hat{a}^\dagger \right] = 1$, it follows that the ket:
$$\hat{a} \, \vert \nu \rangle$$
is also an eigenket of $\hat{n}$, corresponding to an eigenvalue $\nu - 1$. By inductive reasoning, we conclude that $\nu - n, \forall n \in \mathbb{N}$ are also eigenvalues of the number operator.

So, let us suppose that $N + 1 > \nu \ge N \ge 0, \ N \in \mathbb{N}$. This means that $\nu - N - 1 < 0$, which is not allowed (because all the eigenvalues are non-negative!). The only way out is if $\left( \hat{a} \right)^{N + 1} \, \vert \nu \rangle = 0$ (not the ground state $\vert 0 \rangle$, but the zero ket in the Hilbert space!). But, then apply $\left( \hat{a}^\dagger \right)^{N + 1}$ from the left, and use:
===================================================================================
$$\hat{X}_m \equiv \left[ \hat{a}^\dagger, \hat{a}^m \right]$$
$$\hat{X}_{m + 1} = \left[ \hat{a}^\dagger, \hat{a} \, \hat{a}^m \right] = \hat{a} \, \left[ \hat{a}^\dagger, \hat{a}^m \right] + \left[ \hat{a}^\dagger, \hat{a} \right] \, \hat{a}^m = \hat{a} \, \hat{X}_m - \hat{a}^m$$
$$\hat{X}_0 = 0$$
$$\hat{X}_1 = \hat{a} \, 0 - \hat{1} = -\hat{1}$$
$$\hat{X}_2 = \hat{a} \, (-\hat{1}) - \hat{a} = -2 \, \hat{a}$$
...
Lemma:
$$\boxed{\hat{X}_m = -m \, \hat{a}^{m - 1}, \ m \ge 1}$$
Proof:
$$\hat{X}_{m + 1} = \hat{a} \, \left(-m \, \hat{a}^{m - 1} \right) - \hat{a}^{m} = -(m + 1) \, \hat{a}^{m}$$
By Principle of Mathematical Induction, Q.E.D.
===================================================================================

Then, we have:
$$\begin{array}{lcl} \left(\hat{a}^\dagger\right)^{N + 1} \, \hat{a}^{N + 1} \, \vert \nu \rangle & = & 0 \\ \left( \hat{a}^\dagger \right)^N \, \hat{a}^\dagger \, \hat{a}^{N + 1} \, \vert \nu \rangle & = & 0 \\ \left( \hat{a}^\dagger \right)^N \, \left( \hat{a}^{N + 1} \, \hat{a}^\dagger + X_{N + 1} \right) \, \vert \nu \rangle & = & 0 \\ \left[ \left( \hat{a}^\dagger \right)^N \, \hat{a}^{N} \, \hat{a} \, \hat{a}^\dagger - (N + 1) \, \left( \hat{a}^\dagger\right )^{N} \, \hat{a}^{N} \right] \, \vert \nu \rangle & = & 0 \\ \left[ \left( \hat{a}^\dagger \right)^N \, \hat{a}^{N} \left( \hat{n} + 1 \right) - (N + 1) \, \left( \hat{a}^\dagger \right)^N \, \hat{a}^{N} \right] \, \vert \nu \rangle & = & 0 \\ \left( \nu - N \right) \, \left( \hat{a}^\dagger \right)^N \, \hat{a}^{N} \, \vert \nu \rangle & = & 0 \\ \ldots & & \\ \left( \nu - N \right) \, \left( \nu - N + 1\right) \, \ldots \nu \, \vert \nu \rangle & = & 0 \end{array}$$
Now, from the initial assumption $\vert \nu \rangle \neq 0$ (not a trivial eigenket), and $N \le \nu < N + 1$, we conclude that $\nu = N$.

This means that the only eigenvalues of the number operator $\hat{n} = \hat{a}^\dagger \, \hat{a}$ are the non-negative integers. Q.E.D.

6. Jan 21, 2012

### jewbinson

LOL

I meant to leave the question open so that OP could answer it himself... oh well

7. Jan 23, 2012

### aimforclarity

Thanks a lot!