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I was wondering what the proof was that a ladder operator will generate all the eigenenergies for a system. e.g. the QM harmonic oscillator.

So after manipulating the ladder operator we get;

($a$ is ladder operator $a^\dagger$ conjugate) , $|E>$ eigenstate ket, $H$ hamiltonian

$$H= \hbar \omega( 1/2 + a^\dagger a )$$

$$< E|H|E > =E= \hbar \omega(1/2+ < E|a^\dagger a|E > )=\hbar \omega(1/2+(a |E >)^2)$$

as Hamiltonian positive definite, and energy states positive minimum value of energy eigenvalue is when second term is zero. so $E(ground)=(1/2)\hbar \omega$

Also $H a^\dagger |E>=(E+\hbar \omega)a^\dagger|E>$, so on states operated on by $a^\dagger$ the states are still eigenfunctions of the Hamiltonian but have energy of $\hbar \omega$ extra. We can repeat this to find states which have energy $(1/2+n)\hbar \omega$. But what is the proof that these numbers covers all eigenvalues? Could you not have states in between like $(n/2+1/2)$, $(n/3+1/2)$ etc?

Qmechanic
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  • From that same argument it follows that the number n (the eigenvalue of a-dagger a, which at this moment you don't yet assume is an integer), cannot be negative. But applying the lowering operator subtracts 1 from that eigenvalue, so what's to stop you from constructing an eigenstate with a negative eigenvalue? The only thing that can get in the way is if you hit zero on the way, as that will prevent you from normalizing the state. – Count Iblis May 03 '15 at 02:50
  • But if there was a negative eigenvalue the Hamiltonian would not be positive definite I think. The hamiltonian is positive definite as it has only squared terms in it. –  May 03 '15 at 02:53
  • Right, so that cannot happen, therefore n must be an integer to make sure that applying the lowering operator repeatedly will eventually lead to the state |0> where the land mine is located that will stop intrusions into negative n territory. – Count Iblis May 03 '15 at 03:04
  • To put it differently, the lowering operator reduces the eigenvalue by 1 (in units of hbar omega). Now, if |X> is some arbitrary state, then the Hilbert space vector |psi> = a^k|X> for arbitrary k will be well defined. It may not define a physical state if the norm equals zero and turns out to be the null vector. But what cannot happen is that the squared norm of |psi> turns out to be negative. If non-integer n is allowed then the state a^k|n> will become a negative norm state for k sufficiently large. – Count Iblis May 03 '15 at 03:28
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    This question (v2) about the possible spectrum of a SHO is answered in many textbooks. See e.g. this Phys.SE post and link therein. – Qmechanic May 03 '15 at 07:28

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The Schrodinger equation with simple harmonic oscillator potential $V = \frac{1}{2}m\omega^2x^2$

$$-\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2} + \frac{1}{2}m\omega^2x^2\Psi = E\Psi$$

can have solutions of the form

$$\Psi = ke^{ax^2}$$

since

$$-\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2} + \frac{1}{2}m\omega^2x^2\Psi$$ $$=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}(ke^{ax^2}) + \frac{1}{2}m\omega^2x^2(ke^{ax^2})$$ $$=-\frac{\hbar^2k}{2m}\frac{d}{dx}(2axe^{ax^2}) + \frac{1}{2}m\omega^2x^2(ke^{ax^2})$$ $$=-\frac{\hbar^2ka}{m}\frac{d}{dx}(xe^{ax^2}) + \frac{1}{2}m\omega^2x^2(ke^{ax^2})$$ $$=-\frac{\hbar^2ka}{m}(x(2axe^{ax^2}) + e^{ax^2}) + \frac{1}{2}m\omega^2x^2(ke^{ax^2})$$ $$=(-\frac{2\hbar^2a^2x^2}{m} - \frac{\hbar^2 ka}{m}+\frac{1}{2}m\omega^2x^2)ke^{ax^2}$$ $$=E\Psi $$

By physical considerations, $$\Psi_0 = k_0e^{a_0x^2} = (\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}e^{-\frac{m\omega}{2\hbar}x^2}$$

is a fundamental $\Psi$ with the lowest physical $E$.

Any pairs $\Psi_i, \Psi_j$ having the form

$$\Psi_i = \kappa (h\frac{d}{dx}\Psi_j + m\omega x\Psi_j)$$ $$= \kappa (h\frac{d}{dx}(ke^{ax^2}) + m\omega x(ke^{ax^2}))$$ $$= \kappa (2axhke^{ax^2} + m\omega xke^{ax^2})$$ $$= (2\kappa axh + \kappa m\omega x)ke^{ax^2}$$ $$=\bar k xe^{ax^2}$$

can generate infinitely many other solutions, due to the property $\frac{d}{dx} (e^{f(x)}) = k e^{f(x)}$, to

$$-\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2} + \frac{1}{2}m\omega^2x^2\Psi = E\Psi$$

It remains to be shown that all possible solutions are covered by sequentially generating $\Psi_i$ from $\Psi_0$.

James
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