# I Creation operator and Wavefunction relationship

Tags:
1. Jul 4, 2017

### DeathbyGreen

Hello,

I've noticed that some professors will jump between second quantized creation/annihilation operators and wavefunctions rather easily. For instance $\Psi_k \propto c_k + ac_k^{\dagger}$ with "a" some constant (complex possibly). I'm fairly familiar with the second quantized notation, and from my understanding the concept of wavefunctions is essentially abandoned. The creation/annihilation operators simply add a particle into a state in the number basis. So what would be the way to "translate" some single particle wavefunciton $\Psi_k$ into second quantization, and what would the method/reasoning be?

Last edited: Jul 4, 2017
2. Jul 4, 2017

### strangerep

We'd have to see the full context that you're quoting. Sometimes a $\Psi$ like that is indeed an operator, not a state.

3. Jul 4, 2017

### DeathbyGreen

I see what you're saying; in this instance the $\Psi(x)$ is not a field operator, but a single particle wavefunction, like in the case $H\Psi = \Psi E$. Is it legitimate to say that $\Psi(x)$ corresponds to a superposition of $c_k$'s and $c^{\dagger}_k$'s?

4. Jul 4, 2017

### strangerep

... in which case writing $\Psi_k \propto c_k + ac_k^\dagger$ is nonsense (assuming the $c$'s are a/c operators).

5. Jul 5, 2017

### stevendaryl

Staff Emeritus
Quantum field theory can be interpreted as many-particle quantum mechanics where the number of particles is allowed to change. But what's a little confusing is the fact that the expressions $\psi(\vec{r})$ and $\psi^\dagger(\vec{r})$ don't refer to wave functions, but to operators. There is a "wave function" associated with any state with a definite number of particles, although most treatments of quantum field theory don't talk about wave functions much. But it's definitely not correct to associate a wave function with a linear combination of creation and annihilation operators; a wave function is a complex number, not an operator.

6. Jul 5, 2017

### vanhees71

Well, that's why one should mark what's an operator and what's a c-number wave function, i.e., $\hat{\psi}(\vec{r})$ (field operator) vs. $\psi(\vec{r})$ (complex-valued function).

7. Jul 5, 2017

### Jilang

I would see it as a shorthand that omits the vacuum state from the RHS. A bit sloppy but understandable.