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

[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.

Homework Statement

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, $a_+$.
$a_+ = \frac{1}{\sqrt{2 \hbar m \omega}} (-ip + m\omega x)$

Homework Equations

$\left< f | \hat{Q} g \right> = \left< \hat{Q} f | g \right>$

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,
$(a_+)^\dagger = -\frac{1}{\sqrt{2 \hbar m \omega}}(-ip + m \omega x) = -a_+$.

 

[DrDu]

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

 

[BvU]

http://www.physicspages.com/2012/09/05/hermitian-conjugate-of-an-operator/

Hermitian conjugate of any real constant is just the negative of that constant

Hermitian conjugate of any imaginary constant is just the negative of that constant

Good luck with your exam !

 

[anlon]

http://www.physicspages.com/2012/09/05/hermitian-conjugate-of-an-operator/Good luck with your exam !

Ah, so it would be $(a_+)^\dagger = \frac{1}{\sqrt{2 \hbar m \omega}}(ip + m \omega x)$
Thanks, I'll need it.

 

[BvU]

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