• I
• spaghetti3451
In summary: So we have shown that the relation$$\exp(-iHt)\exp(a^{\dagger})\exp(iHt)|0\rangle=\exp(a^{\dagger}e^{-i\omega t})|0\rangle\exp(i\omega t/2)$$holds for the ladder operators of a simple harmonic oscillator.
spaghetti3451
The ladder operators of a simple harmonic oscillator which obey

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

---

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?

As always, try Taylor expansion.

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?

failexam said:
##\exp(-iHt)\exp\left( ca^{\dagger}\right)\exp(iHt)|0,0\rangle##

##= \exp\left( c[a^{\dagger},H]\right)|0,0\rangle##
I don't understand how you go from the first line to the second.

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

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.

Demystifier
failexam said:
The ladder operators of a simple harmonic oscillator which obey

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

---

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?

Is this a homework problem?

stevendaryl said:
Is this a homework problem?

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

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##

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 $(a^\dagger)^n$:

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

vanhees71 and PeroK

## 1. What are harmonic oscillator ladder operators?

Harmonic oscillator ladder operators are mathematical operators that act on the quantum states of a harmonic oscillator system. They are used to raise or lower the energy level of the oscillator and are essential for understanding the behavior of this type of system in quantum mechanics.

## 2. How do harmonic oscillator ladder operators work?

Harmonic oscillator ladder operators work by applying mathematical operations on the quantum states of the system. The raising operator increases the energy level by a fixed amount, while the lowering operator decreases it. These operators follow specific commutation relations that determine their behavior.

## 3. What is the significance of harmonic oscillator ladder operators?

Harmonic oscillator ladder operators play a crucial role in quantum mechanics, particularly in the study of the quantum harmonic oscillator. They allow us to calculate the energy levels of the oscillator and understand its behavior, such as the quantization of energy and the concept of zero-point energy.

## 4. How are harmonic oscillator ladder operators related to the Heisenberg uncertainty principle?

Harmonic oscillator ladder operators are related to the Heisenberg uncertainty principle in that they represent the observable quantities of position and momentum for the harmonic oscillator. These operators do not commute, and the uncertainty principle states that the product of their uncertainties cannot be smaller than a certain value.

## 5. Can harmonic oscillator ladder operators be generalized to other systems?

Yes, harmonic oscillator ladder operators can be generalized to other systems. They are a type of creation and annihilation operators, which are commonly used in quantum mechanics to study the energy levels and behavior of various systems. These operators follow similar commutation relations and can be applied to other types of oscillators or systems with discrete energy levels.

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