简谐振子能量本征态下⟨n|(a+a†)^k|m⟩的便捷恒等式问询

Great question! You're already off to a strong start by working through small $k$ cases—generalizing this to arbitrary natural numbers $k$ does have a clean closed-form identity, which we can derive using the binomial theorem and standard ladder operator properties.

Key Background #

First, recall the core actions of the annihilation ($a$) and creation ($a^\dagger$) operators on harmonic oscillator energy eigenstates $|m\rangle$:

• $a|m\rangle = \sqrt{m}|m-1\rangle$ (for $m \geq 1$; $a|0\rangle = 0$)
• $a^\dagger|m\rangle = \sqrt{m+1}|m+1\rangle$

We start by expanding $(a+a^\dagger)k$ via the binomial theorem:
$$(a+a^\dagger)k = \sum_{p=0}^k \binom{k}{p} a^{k-p} (a^\dagger)p$$

Closed-Form Matrix Element #

The matrix element $\langle n | (a+a^\dagger)k | m \rangle$ is non-zero if and only if:

1. $|n - m| \leq k$ (you can't jump more than $k$ energy levels in $k$ ladder operator steps)
2. $k + n - m$ is an even integer (since each $a$ lowers the state by 1, each $a^\dagger$ raises it by 1; the total level change $n - m = 2p - k$, so $2p = k + n - m$ must be even)

When these conditions are satisfied, the matrix element is given by:
$$\langle n | (a+a^\dagger)k | m \rangle = \binom{k}{\frac{k + n - m}{2}} \cdot \frac{\left( \frac{n + m + k}{2} \right)!}{\sqrt{n! , m!}}$$

Verification with Small $k$ #

Let's confirm this matches your existing calculations:

• $k=1$: For $n=m+1$, $p=(1 + (m+1)-m)/2=1$, $\binom{1}{1}=1$, and $\frac{(m+1)!}{\sqrt{(m+1)!m!}} = \sqrt{m+1}$ (matches $\langle m+1|a^\dagger|m\rangle$). For $n=m-1$, $p=0$, $\binom{1}{0}=1$, and $\frac{m!}{\sqrt{(m-1)!m!}} = \sqrt{m}$ (matches $\langle m-1|a|m\rangle$).
• $k=2$: For $n=m$, $p=(2 + m - m)/2=1$, $\binom{2}{1}=2$, and $\frac{(m+1)!}{\sqrt{m!m!}} = m+1$, giving $2(m+1)$ (matches $\langle m|(a+a^\dagger)2|m\rangle = 2(m+1)$).

Derivation Breakdown #

Here's how we get to the closed-form:

1. For each term in the binomial expansion, compute $\langle n | a^{k-p} (a^\dagger)p | m \rangle$.
2. Apply $a^\dagger$ $p$ times to $|m\rangle$: $(a^\dagger)p|m\rangle = \sqrt{\frac{(m+p)!}{m!}} |m+p\rangle$.
3. Apply $a$ $k-p$ times to $|m+p\rangle$: $a^{k-p}|m+p\rangle = \sqrt{\frac{(m+p)!}{n!}} |n\rangle$ (since we require $m+p - (k-p) = n$, which rearranges to $p = \frac{k + n - m}{2}$).
4. The inner product $\langle n | |n\rangle = 1$, so multiply the constants and binomial coefficient to get the final result.

Alternative Recurrence Relation #

If you prefer a recursive approach (useful for numerical work or checking cases), use the linearity of the operator:
$$\langle n | (a+a^\dagger)k | m \rangle = \sqrt{m} \langle n | (a+a^\dagger){k-1} | m-1 \rangle + \sqrt{m+1} \langle n | (a+a^\dagger){k-1} | m+1 \rangle$$
Base cases:

• $\langle n | (a+a^\dagger)0 | m \rangle = \delta_{n,m}$ (identity operator)
• $\langle n | (a+a^\dagger)1 | m \rangle = \sqrt{m}\delta_{n,m-1} + \sqrt{m+1}\delta_{n,m+1}$

内容的提问来源于stack exchange，提问作者Diffycue