# I Harmonic oscillator ladder operators

1. Dec 2, 2016

### spaghetti3451

The ladder operators of a simple harmonic oscillator which obey

$$[H,a^{\dagger}]=\hbar\omega\ a^{\dagger}$$.

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I would like to see a proof of the relation

$$\exp(-iHt)\exp(a^{\dagger})\exp(iHt)|0\rangle=\exp(a^{\dagger}e^{-i\omega t})|0\rangle\exp(i\omega t/2).$$

Thoughts?

2. Dec 2, 2016

### Staff: Mentor

As always, try Taylor expansion.

3. Dec 2, 2016

### spaghetti3451

I did this.

$\exp(-iHt)\exp\left( ca^{\dagger}\right)\exp(iHt)|0,0\rangle$

$= \exp\left( c[a^{\dagger},H]\right)|0,0\rangle$

$= \exp\left( ca^{\dagger}\exp(-i\omega t)\right)\exp(i\omega t)|0,0\rangle$

Where have I gone wrong?

4. Dec 2, 2016

### Staff: Mentor

I don't understand how you go from the first line to the second.

5. Dec 2, 2016

### spaghetti3451

I used Heisenberg's equation of motion for the ladder operator in the Heisenberg picture.

6. Dec 2, 2016

### vanhees71

The Heisenberg equations of motion are a good idea, but it's perhaps easier to use the Trotter product formula,

http://theory.gsi.de/~vanhees/faq/quant/node99.html

The link is in German, but the formula density should be high enough to get what you need. Note that in your case you can resum the resulting series explicitly.

7. Dec 2, 2016

8. Dec 2, 2016

### stevendaryl

Staff Emeritus
Is this a homework problem?

9. Dec 2, 2016

### spaghetti3451

No, it's not! This is from a paper that I am reading.

10. Dec 2, 2016

### spaghetti3451

Sorry, I made a mistake. This is what I am trying to prove.

$\exp(-iHt)\exp(a^{\dagger})\exp(iHt)|0\rangle=\exp(a^{\dagger}e^{-i\omega t})|0\rangle$

11. Dec 2, 2016

### stevendaryl

Staff Emeritus
Okay, so it might be slightly easier to work with the "number" operator $N$, which is related to $H$ via:

$H = (N+1/2) \omega$ (in units where $\hbar = 1$)

In terms of the number operator, $[N, a^\dagger] = a^\dagger$.

So we want to show that:

$e^{-i (N+1/2) \omega t} e^{a^\dagger} e^{+i (N+1/2) \omega t} |0\rangle = exp(a^\dagger e^{-i\omega t}) |0\rangle$

Since $e^{-i/2 \omega t}$ and $e^{+i/2 \omega t}$ are just numbers, not operators, they commute, and they combine to form 1. So we need to show:

$e^{-i N \omega t} e^{a^\dagger} e^{+i N \omega t} |0\rangle = exp(a^\dagger e^{-i\omega t}) |0\rangle$

Here, we can write $e^{i N \omega t} = 1 + i N\omega t + 1/2 (i N \omega t)^2 + ...$. When this acts to the right on $|0\rangle$, we just get $|0\rangle$, because $N |0\rangle = 0$. So we want to show:

$e^{-i N \omega t} e^{a^\dagger} |0\rangle = exp(a^\dagger e^{-i\omega t}) |0\rangle$

Now, we write $e^{a^\dagger}$ as $\sum_n (a^\dagger)^n/n!$

We can prove (by induction) that if $|n\rangle$ is the normalized state in which $N |n\rangle = n |n \rangle$, then $(a^\dagger)^n |0\rangle = \sqrt{n!} |n\rangle$

So we have, on the left-hand side:

$e^{-i N \omega t} \sum_n \frac{\sqrt{n!}}{n!} |n\rangle$

Now, you can bring the first exponent inside the sum to get:

$\sum_n \frac{\sqrt{n!}}{n!} e^{-i N \omega t} |n\rangle$

Since $N |n\rangle = n |n\rangle$, this just becomes:

$\sum_n \frac{\sqrt{n!}}{n!} e^{-i n \omega t} |n\rangle$
$= \sum_n \frac{\sqrt{n!}}{n!} (e^{-i \omega t})^n |n\rangle$

Now, we can undo the application of $(a^\dagger)^n$:

$\sum_n \frac{\sqrt{n!}}{n!} (e^{-i \omega t})^n |n\rangle = \sum_n (e^{-i \omega t})^n (a^\dagger)^n/n!|0\rangle$
$= \sum_n (a^\dagger e^{-i \omega t})^n/n!|0\rangle$
$= exp(a^\dagger e^{-i \omega t})|0\rangle$