- #1

Luke Tan

- 29

- 2

- 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)

- Relevant 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?

$$a^\dagger \lvert n\rangle = C_{n+1}\lvert n+1 \rangle$$

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?