Question about creation and annihilation operators?

[21joanna12]

Hello! I am reading about the creation and annihilation operators and I don't get how you find the creation operator from the annihilation one. The creation one is

\hat{a}=\sqrt{\frac{m \omega}{2 \hbar}}\left( \hat{x}+\frac{i \hat{p}}{m \omega}\right)

and the annihilation operator is \hat{a}\dagger =\sqrt{\frac{m \omega}{2 \hbar}}\left( \hat{x}-\frac{i \hat{p}}{m \omega}\right)

I don't understand how taking the complex conjugate an transpose leads to the minus sign. I thought that \hat{x}=x and \hat{p}=i\hbar\frac{\partial}{\partial x} so \hat{x}\dagger=\hat{x}=x but
\hat{p}\dagger=-i\hbar\frac{\partial}{\partial x}=-\hat{p} which will cancel with the minus sign from complex conjugating the i, so that \hat{a}=\hat{a}\dagger.

I'm sure that this mistake is due to my unfamiliarity with the algebra of operators. I would appreciate it if someone could say which of my assumptions is wrong! My hunch is that it has something to do with taking the dagger of \left( \hat{x}+\frac{i \hat{p}}{m \omega}\right) and this not being equal to taking the dagger of the individual operators...

Thanks in advance!

 

[CAF123]

The problem is where you computed $p^{\dagger}$. Since this is an operator corresponding to an observable (momentum), you know that it must be hermitian so that $p^{\dagger} = p$. Therefore your error is in the computation of $p^{\dagger}$. Find first how the operator $d/dx$ transforms under a hermitian conjugate.

 

[stevendaryl]

I don't understand how taking the complex conjugate an transpose leads to the minus sign. I thought that \hat{x}=x and \hat{p}=i\hbar\frac{\partial}{\partial x} so \hat{x}\dagger=\hat{x}=x but
\hat{p}\dagger=-i\hbar\frac{\partial}{\partial x}=-\hat{p} which will cancel with the minus sign from complex conjugating the i, so that \hat{a}=\hat{a}\dagger.

For an operator, \hat{O}, the definition of \hat{O}^\dagger is a little technical: If you have two wave functions \psi and \phi, then according to Dirac notation:

\langle \phi | \psi \rangle = \int \phi(x)^* \psi(x) dx

Then the meaning of \hat{O}^\dagger is given by:

\langle \phi | \hat{O} \psi \rangle = \langle \hat{O}^\dagger \phi | \psi\rangle

In the case of a "hermitian" operator, \hat{O}^\dagger = \hat{O}, so we can just write:

\langle \phi | \hat{O} | \psi \rangle

and not worry about whether \hat{O} is acting to the right, on \psi, or acting to the left, on \phi.

In the special case where \hat{O} is the derivative operator, \frac{d}{dx}, you can prove^\dagger:

\langle \phi | \frac{d}{dx} \psi \rangle = \langle (- \frac{d}{dx} \phi) | \psi \rangle

So (\frac{d}{dx})^\dagger = - \frac{d}{dx}

^\daggerThis is only true when \phi and \psi are well-behaved wave functions. You prove this by integration by parts, and reasoning that \phi and \psi go to zero as x \rightarrow \pm \infty.

 

[21joanna12]

For an operator, \hat{O}, the definition of \hat{O}^\dagger is a little technical: If you have two wave functions \psi and \phi, then according to Dirac notation:

\langle \phi | \psi \rangle = \int \phi(x)^* \psi(x) dx

Then the meaning of \hat{O}^\dagger is given by:

\langle \phi | \hat{O} \psi \rangle = \langle \hat{O}^\dagger \phi | \psi\rangle

In the case of a "hermitian" operator, \hat{O}^\dagger = \hat{O}, so we can just write:

\langle \phi | \hat{O} | \psi \rangle

and not worry about whether \hat{O} is acting to the right, on \psi, or acting to the left, on \phi.

In the special case where \hat{O} is the derivative operator, \frac{d}{dx}, you can prove^\dagger:

\langle \phi | \frac{d}{dx} \psi \rangle = \langle (- \frac{d}{dx} \phi) | \psi \rangle

So (\frac{d}{dx})^\dagger = - \frac{d}{dx}

^\daggerThis is only true when \phi and \psi are well-behaved wave functions. You prove this by integration by parts, and reasoning that \phi and \psi go to zero as x \rightarrow \pm \infty.

Brilliant! Thank you both! I see now how \hat{p}\dagger=\hat{p}, but I'm also wondering whether you can distribute the 'dagger' over the bracket \left(\hat{x}+\frac{i\hat{p}}{m\omega}\right) by simply taking the 'dagger' of the components and i? Or am I thinking about this wrong?

Thank you! :)

 

[stevendaryl]

Brilliant! Thank you both! I see now how \hat{p}\dagger=\hat{p}, but I'm also wondering whether you can distribute the 'dagger' over the bracket \left(\hat{x}+\frac{i\hat{p}}{m\omega}\right) by simply taking the 'dagger' of the components and i? Or am I thinking about this wrong?

Thank you! :)

Yes, the \dagger operation distributes.

 

[stevendaryl]

Yes, the \dagger operation distributes.

Technically, (AB + C)^\dagger = B^\dagger A^\dagger + C^\dagger

 

[dextercioby]

The harmonic oscillator (in 1D) is the simplest possible example of how functional analysis makes the heart of quantum mechanics. A nice propery which enables us to use ladder operators is the fact that \mathcal{S}(\mathbb{R}) the Schwartz test functions space is a common dense everywhere domain of essential self-adjointness for both x and p. Hence it makes sense to consider the adjoint of a = x + \alpha p , where \alpha is a complex constant (number, not operator).

a^{\dagger} = \left(x + \alpha p \right)^{\dagger} \supseteq x^{\dagger} + (\alpha p)^{\dagger} \supseteq x +\alpha^{*} p

Since i^{*} = -i, that's how you end up with a minus before p. If the operators are restricted to \mathcal{S}(\mathbb{R}) from \mathcal{L}^2(\mathbb{R}), you're safe to use the equality sign.