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

• anlon
In summary: Overall, you did a great job summarizing the conversation about Hermitian conjugates and finding their values for different operators. In summary, the Hermitian conjugate of an operator is equal to itself, except for the case of imaginary constants where it is equal to the negative of that constant. This is true for the operators x, i, and d/dx. For the harmonic oscillator raising operator, a+, its Hermitian conjugate is equal to the negative of itself. Good luck on your exam!
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_+##.

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

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

anlon said:
Hermitian conjugate of any real constant is just the negative of that constant

anlon said:
Hermitian conjugate of any imaginary constant is just the negative of that constant

Good luck with your exam !

BvU said:
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.

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

## 1. What is a Hermitian conjugate?

A Hermitian conjugate is the complex conjugate transpose of a matrix or operator. This means that for a given matrix A, its Hermitian conjugate is found by taking the complex conjugate of each element and then transposing the matrix.

## 2. How do I find the Hermitian conjugate of x?

The Hermitian conjugate of x is simply its complex conjugate, as x is a real number. The complex conjugate of a real number is the number itself, so the Hermitian conjugate of x is just x.

## 3. How do I find the Hermitian conjugate of i?

The Hermitian conjugate of i is -i. This is because i is a complex number with a real part of 0 and an imaginary part of 1, so its complex conjugate is -i.

## 4. How do I find the Hermitian conjugate of d/dx?

The Hermitian conjugate of d/dx is -d/dx. This is because the Hermitian conjugate of a derivative operator is found by taking the negative of the derivative operator.

## 5. Why is finding the Hermitian conjugate important?

Finding the Hermitian conjugate is important in quantum mechanics and other areas of physics where complex numbers are used. Hermitian operators and matrices have special properties that make them useful in these applications, and the Hermitian conjugate allows us to calculate these properties and relationships.

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