# Find the Hermitian conjugates: $x$, $i$,$\frac{d}{dx}$

1. Dec 7, 2016

### anlon

Just doing some studying before my final exam later today. I think I've got this question right but wanted to make sure since the problem is from the international edition of my textbook, so I can't find the solutions for that edition online.
1. The problem statement, all variables and given/known data
The Hermitian conjugate (or adjoint) of an operator $\hat{Q}$ is the operator $\hat{Q}$ such that $$\left<f | \hat{Q} g \right> = \left<\hat{Q} | g\right>, \ \text{for all }f \text{ and }g$$
(A Hermitian operator, then, is equal to its Hermitian conjugate: $\hat{Q} = \hat{Q}$.)
a. Find the Hermitian conjugates of $x$, $i$, and $\frac{d}{dx}$.
b. Construct the Hermitian conjugate of the harmonic oscillator raising operator, $a_+$.
$a_+ = \frac{1}{\sqrt{2 \hbar m \omega}} (-ip + m\omega x)$

2. Relevant equations
$\left< f | \hat{Q} g \right> = \left< \hat{Q} f | g \right>$

3. The attempt at a solution
For $x$: $\left< f | x g \right> = \int_{-\infty}^{\infty} f^{*} x g dx$
Also, $\left< x f | g \right> = \int_{-\infty}^{\infty} (x f)^{*} g dx$
If $x$ is real, then $x^{*} = x$
So $\int_{-\infty}^{\infty} (x f)^{*} g dx = \int_{-\infty}^{\infty} x f^{*} g dx = \int_{-\infty}^{\infty} f^{*} x g dx$
So $x$ is a Hermitian operator; its Hermitian conjugate is itself.

For $i$: $\left< f | i g \right> = \int_{-\infty}^{\infty} f^{*} i g dx$
Also, $\left< i f | g \right> = \int_{-\infty}^{\infty} (i f)^{*} g dx = \int_{-\infty}^{\infty} f^{*} (-i) g dx$
So $\left< f | i g \right> = \left< (-i) f | g \right>$ which means that the Hermitian conjugate of $i$ is $-i$.

For $\frac{d}{dx}$: (this is the one I'm unsure about)
$\left< f | \frac{d}{dx} g \right> = \int_{-\infty}^{\infty} f^{*} \frac{dg}{dx} dx$
Integrate by parts: let $u = f^{*}$ and $dv = \frac{dg}{dx} dx$.
Then $du = \frac{df}{dx}^{*} dx$ and $v = g$
Integrating by parts, $\int_{-\infty}^{\infty} f^{*} \frac{dg}{dx} dx = f^{*} g |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{df}{dx}^{*} g dx$
The term on the left goes to zero since f(x) and g(x) are square integrable, so they must be zero at either extreme.
So the Hermitian conjugate of $\frac{d}{dx}$ is $-\frac{d}{dx}$.

For $a_+$: (writing this quickly because I have to get back to studying, and formatting equations takes a while)
$\left< f | a_+ g \right> = \left< f | \frac{1}{\sqrt{2 \hbar m \omega}}(-ip + m \omega x) g \right>$
$p$ and $x$ are Hermitian, and the Hermitian conjugate of any constant is just the negative of that constant. So this should be equivalent to
$\left< -\frac{1}{\sqrt{2 \hbar m \omega}} ( -i p + m \omega x) f | g \right>$
Therefore,
$(a_+)^\dagger = -\frac{1}{\sqrt{2 \hbar m \omega}}(-ip + m \omega x) = -a_+$.

2. Dec 7, 2016

### DrDu

The first three proofs are perfect, but you made a mistake with a+.

3. Dec 7, 2016

4. Dec 7, 2016

5. Dec 7, 2016

### BvU

Yes, $(a_+)^\dagger = a_-$ and vice versa