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anlon
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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.
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)##
##\left< f | \hat{Q} g \right> = \left< \hat{Q} f | g \right>##
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_+##.
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_+##.
