# Creation Operator

#### Luke Tan

Homework Statement
Show that $a^\dagger\lvert n \rangle = \sqrt{n+1}\lvert n+1 \rangle$
(From Shankar, Principles of Quantum Mechanics, Chapter 8 - The Harmonic Oscillator)
Homework Equations
$\hat{H}+\frac{1}{2}=aa^\dagger=a^\dagger a$
$\hat{H}\lvert n \rangle = (n+\frac{1}{2})\lvert n \rangle$
I have written the equation, with an unknown constant
$$a^\dagger \lvert n\rangle = C_{n+1}\lvert n+1 \rangle$$
I then take the adjoint to get
$$\langle n \rvert a = \langle n+1 \rvert C_{n+1}^\text{*}$$
I then multiply them to get
$$\langle n \rvert aa^\dagger \lvert n \rangle = \langle n+1 \rvert n+1 \rangle |C_{n+1}|^2$$
On the left hand side, since $aa^\dagger = \hat{H}-\frac{1}{2}$, the expression just simplifies to $\langle n \rvert n \lvert n \rangle$. On the right hand side, since $\lvert n+1 \rangle$ is a normalized state, it just simplifies to $|C_{n+1}|^2$. Thus, we arrive at
$$\langle n \rvert n \lvert n \rangle = |C_{n+1}|^2$$
$$n = |C_{n+1}|^2$$
$$C_{n+1}=\sqrt{n}$$.
Thus,
$$a^\dagger \lvert n \rangle = \sqrt{n} \lvert n+1 \rangle$$
Which is wrong.

I can't see where i went wrong. Can someone help?

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#### jambaugh

Science Advisor
Gold Member
First error I see is where you wrote: $aa^\dagger = a^\dagger a$ their commutator is not 0.

#### Luke Tan

First error I see is where you wrote: $aa^\dagger = a^\dagger a$ their commutator is not 0.
Would it make any difference?
From my equation

$$\langle n \rvert aa^\dagger \lvert n \rangle = \langle n+1 \rvert n+1 \rangle |C_{n+1}|^2$$

Could i take the adjoint to get
$$\langle n \rvert a^\dagger a \lvert n \rangle = \langle n+1 \rvert n+1 \rangle |C_{n+1}|^2$$

which would result in the same expression?

#### hutchphd

The number operator is self-adjoint (conjugate transpose). Yes it makes a big difference!!

#### Orodruin

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
2018 Award
Could i take the adjoint to get
That is wrong. The adjoint of $a a^\dagger$ is $aa^\dagger$, not $a^\dagger a$. Note that $(ab)^\dagger = b^\dagger a^\dagger$ for general a and b.

#### Luke Tan

Oh i got the answer

We start from the original equation
$$a^\dagger \lvert n\rangle = C_{n+1}\lvert n+1 \rangle$$

We then take the adjoint to get
$$\langle n \rvert a = \langle n+1 \rvert C_{n+1}^\text{*}$$

Which we then combine to arrive at
$$\langle n \rvert aa^\dagger \lvert n \rangle = \langle n+1 \rvert n+1 \rangle |C_{n+1}|^2$$

Since we know the commutator $[a,a^\dagger]=aa^\dagger-a^\dagger a=1$, we can then derive $aa^\dagger = 1 + a^\dagger a$. We then substitute this to get

$$\langle n \rvert 1 + a^\dagger a \lvert n \rangle = \langle n+1 \rvert n+1 \rangle |C_{n+1}|^2$$

Which then simplifies to

$$\langle n \rvert \hat{H}+\frac{1}{2} \lvert n \rangle = |C_{n+1}|^2$$

Which evaluates to $\langle n \rvert n+1 \lvert n \rangle = |C_{n+1}|^2$

Thus, we get $|C_{n+1}|=\sqrt{n+1}$

Thanks everyone for the help!

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