Second quantization and creation/annih. operators

  • #1

Main Question or Discussion Point

I'd like some help with something in this introduction to second quantization ... http://yclept.ucdavis.edu/course/242/Class.html

They start by considering two operators ##a## and ##a^\dagger## such that ##[a,a^\dagger]=1## and they are adjoint to each other. Also introduced is a state ##|\alpha\rangle## that is an eigenvector of the (Hermitian) operator ##a^\dagger a##, so that
##a^\dagger a |\alpha\rangle = \alpha |\alpha\rangle##.

They go on to show that the state ##a|\alpha\rangle## is an eigenstate of ##a^\dagger a## with eignvalue ##\alpha-1##, and ##a^\dagger|\alpha\rangle## is an eigenstate of ##a^\dagger a## with eigenvalue ##\alpha-1##.

Up to this point I managed to follow the plot reasonably well.

But now they say that the above "implies" that

##|\alpha-1\rangle=\frac 1{\sqrt\alpha}\alpha|\alpha\rangle##
and
##|\alpha+1\rangle=\frac 1{\sqrt{\alpha+1}}\alpha|\alpha\rangle##

which I am absulutely unable to understand. How do they get this result? where do the square roots come from? Why is there an asymmetry in the square roots?
 

Answers and Replies

  • #2
Ssnow
Gold Member
509
149
The square root come from the normalization, in order to have that the number of particle in the state ##\alpha##, so ##N_{\alpha}=a^{\dagger}a=n_{\alpha}##. I think that the asymmetry consist in that: in one case you have ##\sqrt{\alpha+1}## and in the other ##\sqrt{\alpha}## and not ##\sqrt{\alpha-1}## as the vector ##|\alpha-1\rangle##...
 
  • #3
Thanks for the reply.. One reason I am confused is that, up to this point, they are talking about ##a## and ##a^\dagger## as two abstract operators with certain assumed properties -- they have not yet brought in any physics. And they seem to be saying that the square roots follow from (are implied by) the discussion so far. So I'm trying to understand how to derive those terms from what precedes.
 
  • #4
954
117
##a^\dagger|\alpha\rangle## is an eigenstate of ##a^\dagger a## with eigenvalue ##\alpha-1##
I'm pretty sure it should be ##\alpha+1##, and that
##|\alpha-1\rangle=\frac 1{\sqrt\alpha}\alpha|\alpha\rangle##
and
##|\alpha+1\rangle=\frac 1{\sqrt{\alpha+1}}\alpha|\alpha\rangle##
should be
##|\alpha-1\rangle=\frac 1{\sqrt\alpha}\hat{a}|\alpha \rangle## and ##|\alpha+1\rangle=\frac 1{\sqrt{\alpha+1}}\hat{a}^{\dagger}|\alpha\rangle##.
This result can be obtained by demanding that the state kets are normalised - for example, we know that ##a^\dagger|\alpha\rangle## is proportional to ##|\alpha+1\rangle##, and we can determine the constant of proportionality by calculating the norm on both sides.
 

Related Threads on Second quantization and creation/annih. operators

Replies
6
Views
2K
  • Last Post
Replies
2
Views
902
  • Last Post
Replies
1
Views
694
Replies
34
Views
9K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
5K
Replies
2
Views
7K
Top