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

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

^$\dagger$This 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.