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Creation Operator

  • Thread starter Luke Tan
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11
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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?
 

jambaugh

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First error I see is where you wrote: [itex]aa^\dagger = a^\dagger a[/itex] their commutator is not 0.
 
11
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First error I see is where you wrote: [itex]aa^\dagger = a^\dagger a[/itex] 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?
 
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The number operator is self-adjoint (conjugate transpose). Yes it makes a big difference!!
 

Orodruin

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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.
 
11
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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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