Ladder operators and the Hermitian adjoint

Question

We define $$\hat{a}=\sqrt{\frac{m \omega}{2 \hbar}}\left(\hat{x}+i \frac{\hat{p}}{m w}\right)$$ $$\hat{a}^{\dagger}=\sqrt{\frac{m \omega}{2 \hbar}}\left(\hat{x}-i \frac{\hat{p}}{m w}\right)$$ Lowering and raising operators respectively. Clearly, these are conjugates in the regular complex scalar sense but we label them and treat them as Hermitian adjoints (complex transpose). Since the differential operator can be represented by an infinitely large matrix, why do we not need to take the transpose part of it into account?

Answer 1

In the number basis of the harmonic oscillator (where the number operator energy is diagonal), your text must (and certainly Wikipedia does) have $$a^\dagger =\begin{pmatrix} 0 & 0 & 0 & \dots & 0 &\dots \\ \sqrt{1} & 0 & 0 & \dots & 0 & \dots\\ 0 & \sqrt{2} & 0 & \dots & 0 & \dots\\ 0 & 0 & \sqrt{3} & \dots & 0 & \dots\\ \vdots & \vdots & \vdots & \ddots & \vdots & \dots\\ 0 & 0 & 0 & \dots & \sqrt{n} &\dots & \\ \vdots & \vdots & \vdots & \vdots & \vdots &\ddots \end{pmatrix} $$ and $$a =\begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \dots & 0 & \dots \\ 0 & 0 & \sqrt{2} & 0 & \dots & 0 & \dots \\ 0 & 0 & 0 & \sqrt{3} & \dots & 0 & \dots \\ 0 & 0 & 0 & 0 & \ddots & \vdots & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \sqrt{n} & \dots \\ 0 & 0 & 0 & 0 & \dots & 0 & \ddots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}, $$

both infinite-dimensional matrices and adjoint (actually transpose) of each other.

Adding them together gives you the hermitian (actually symmetric) $$ \hat x = \sqrt{\frac{\hbar}{2m\omega}} \begin{pmatrix} 0 & \sqrt{1} & 0 & 0 & \dots & 0 & \dots \\ \sqrt{1} & 0 & \sqrt{2} & 0 & \dots & 0 & \dots \\ 0 & \sqrt{2} & 0 & \sqrt{3} & \dots & 0 & \dots \\ 0 & 0 & \sqrt{3} & 0 & \ddots & \vdots & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots & \sqrt{n} & \dots \\ 0 & 0 & 0 & 0 & \sqrt{n} & 0 & \ddots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}, $$ and it should be straightforward for you to construct the likewise hermitian (imaginary antisymmetric) $\hat p$.

Answer 2

Short answer: $x$ and $p$ in quantum mechanics are operators, not numbers. It means that, for example, $\hat{x}|\psi\rangle = |\phi\rangle$ where $|\psi\rangle$ and $|\phi\rangle$ vectors in our Hilbert space (up to normalization). As they are operators that correspond to physical observables, they must be Hermitian operators. The rules of how to take the Hermitian-conjugate of $a$ to get $a^{\dagger}$ follow immediately.

Longer answer: you are right that the usual definition of operators in terms of matrices in quantum mechanic is sometime problematic, especially at undergraduate level. The point is that $x$ and $p$ cannot be defined in terms of matrices of finite size. To see this, it is suffice to note that $\left[ x,p \right] = i\hbar$ is a requirement, yet for any two finite matrices $A$ and $B$, we have ${\rm Tr}\{\left[A, B\right]\} =0$ because the trace is cyclic.

There are ways to write $x$ and $p$ as matrices, but they must be either infinite, or violate this fundamental commutation relation. Now, there is no problem with infinite matrices. In fact, $a$ and $a^{\dagger}$, represented as matrices, must also be infinite (just because they also maintain a commutation relation of $[a,a^{\dagger}]=1$). This comes from the fact that our Hilbert space is infinite $|n\rangle$ with $n$ any non-negative integer.

One can define $x$ and $p$ as finite matrices, with some finite-lattice spacing, in a large box $L$, and then demand that $\langle \psi | [x,p] | \psi \rangle = i\hbar$ for almost the entire Hilbert space, with the states for which this is violated have their support only at the edges of the box, and we are never to use them for any physical observable. This is sometimes done when we want to write $x$ and $p$ on the computer and calculate stuff, and we are of course restricted only to finite matrices.

Answer 3

In the language of a Hilbert space, the Hermitian conjugate $A^{\dagger}$ of an operator $A$ is defined by (neglecting important issues about the respective domains of these operators): $$ \langle A^{\dagger} u, v \rangle = \langle u, A v \rangle \quad \forall\, u, v \in H $$ We can put this in bra-ket notation by writing $$ \langle \xi | \psi \rangle = \langle \phi | \hat{\,A} | \psi \rangle \quad \text{where} \quad |\xi \rangle = \hat{\,A}^{\dagger} |\phi \rangle \quad \quad \forall\,| \phi \rangle, | \psi \rangle \in H $$ So, we wish to show that: $$ \langle \xi | \psi \rangle = \langle \phi | \hat{\,a}_- | \psi \rangle \quad \text{where} \quad |\xi \rangle = \hat{\,a}_+ | \phi \rangle $$ Now we can calculate the left-hand side using the position basis: \begin{align} \langle \xi | \psi \rangle &= \langle \xi| \left[ \int {\rm d}x \, | x \rangle \langle x | \right] | \psi \rangle\\ &= \int {\rm d}x\, \langle \xi | x \rangle \langle x | \psi \rangle\\ &= \int {\rm d}x\, \langle x | \xi \rangle^* \langle x | \psi \rangle\\ &= \int {\rm d}x\, \left[ \langle x | \hat{\,a}_+ | \phi \rangle \right]^* \langle x | \psi \rangle \\ &= \int {\rm d}x\, \left[\langle x | \frac{1}{\sqrt{2m}} \left( \hat{\,p} + i\, m \omega \hat{\,x} \right) | \phi \rangle \right]^* \langle x | \psi \rangle \\ &= \frac{1}{\sqrt{2m}} \int {\rm d}x\, \left[ -i \hbar \frac{\rm d}{{\rm d}x} \langle x| \phi\rangle + i \,m \omega x \, \langle x | \phi \rangle \right]^* \langle x | \psi \rangle \\ &= \frac{1}{\sqrt{2m}} \int {\rm d}x\, \left[ +i \hbar \frac{{\rm d} \langle \phi| x \rangle}{{\rm d}x} \langle x | \psi \rangle - i \, m \omega x \, \langle \phi | x \rangle \langle x | \psi \rangle \right] \\ &= \frac{1}{\sqrt{2m}} \left[ \langle \phi| x\rangle \langle x | \psi \rangle \right]_{-\infty}^{+\infty} + \frac{1}{\sqrt{2m}} \int {\rm d}x\, \left[ -i \hbar \langle \phi | x \rangle \frac{{\rm d} \langle x| \psi \rangle}{{\rm d}x} - i \, m \omega x \, \langle \phi | x \rangle \langle x | \psi \rangle \right] \\ &= \int {\rm d}x\, \langle \phi | x \rangle \langle x | \frac{1}{\sqrt{2m}} \left[ \hat{\,p} - i \, m \omega \, \hat{\, x}\right] |\psi \rangle \\ &= \langle \phi | \left[ \int {\rm d}x\, |x \rangle \langle x| \right] \hat{\,a}_- | \psi \rangle = \langle \phi | \hat{\,a}_- | \psi \rangle \end{align} where we've used integration by parts and the fact that the wavefunction representations of $| \phi\rangle$ and $|\psi\rangle$ must vanish at $\pm \infty$.