Physics Help Forum Ladder operators/Dirac notation

 Quantum Physics Quantum Physics Help Forum

 Apr 17th 2008, 03:13 AM #1 Junior Member   Join Date: Apr 2008 Posts: 4 Ladder operators/Dirac notation I have this problem and the solution is given, but I cannot understand how it was arrived at. <0| (a + a+)^4 |0> = <0| a(a + a+)(a + a+)a+ |0> (apologies for the notation, but a+ represents the raising ladder operator, and a is the lowering operator) I don't understand how you get this step. Secondly, when the 2 middle terms are expanded, you get a^2 and (a+)^2...which equals 0, so we're left with: <0| a( a(a+) + (a+)a )a+ |0> I've seen a similar thing in other examples as well - why should this be ? Thanks for any help.
Apr 17th 2008, 05:52 AM   #2

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,494
 Originally Posted by sathys I have this problem and the solution is given, but I cannot understand how it was arrived at. <0| (a + a+)^4 |0> = <0| a(a + a+)(a + a+)a+ |0> (apologies for the notation, but a+ represents the raising ladder operator, and a is the lowering operator) I don't understand how you get this step.
Note that $\displaystyle a|0> = 0$ and $\displaystyle <0|a^{\dagger} = 0$, so
$\displaystyle <0| \left ( a + a^{\dagger} \right ) ^4|0>$

$\displaystyle = \left ( <0|(a + a^{\dagger}) \right ) \left ( a + a^{\dagger} \right )^2 \left ( (a + a^{\dagger}) |0> \right )$

$\displaystyle = \left ( <0|a + <0|a^{\dagger}) \right ) \left ( a + a^{\dagger} \right )^2 \left ( a|0> + a^{\dagger} |0> \right )$

$\displaystyle = ( <0|a ) \left ( a + a^{\dagger} \right )^2 ( a^{\dagger} |0> )$

$\displaystyle = <0|a \left ( a + a^{\dagger} \right )^2 a^{\dagger} |0>$

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

Apr 17th 2008, 05:57 AM   #3

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,494
 Originally Posted by sathys Secondly, when the 2 middle terms are expanded, you get a^2 and (a+)^2...which equals 0, so we're left with: <0| a( a(a+) + (a+)a )a+ |0> I've seen a similar thing in other examples as well - why should this be ? Thanks for any help.
In general $\displaystyle a^2 + \left ( a^{\dagger} \right )^2 \neq 0$

However when we take them bracketed by the same states:
$\displaystyle <n|a^2|n> = 0$
and
$\displaystyle <n| \left ( a^{\dagger} \right ) ^2 |n> = 0$

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Apr 17th 2008, 09:46 AM #4 Junior Member   Join Date: Apr 2008 Posts: 4 Thanks Dan. I understand most of it, but have a couple of follow-up questions... I'm not completely sure why : $\displaystyle <0|a^{\dagger} = 0$ , is true. My main problem is I'm not very comfortable with manipulating things in Dirac notation (i.e. what you're 'allowed' to do). Do you have a short proof for this? Again, do you have a proof for why the squares are 0, when bracketed by the same states? Now, I'm able to follow your working for the rest of the question, and can vaguely follow the logic to the rest of the question but I would like to run things over with you, just to make sure my reasoning is correct. The next steps are: $\displaystyle <0|a \left (a + a^{\dagger} \right)^2a^{\dagger}|0>$ $\displaystyle = <0|a \left (aa^{\dagger} + a^{\dagger}a \right) a^{\dagger} |0>$ Using the commutator relation $\displaystyle [a,a^{\dagger}]=1$, we get $\displaystyle = <0|a \left (1 + 2a^{\dagger}a \right) a^{\dagger} |0>$ $\displaystyle = <0|aa^{\dagger}|0> + 2<0|aa^{\dagger} \left(a^{\dagger}a + 1 \right) |0>$ This is where I wanted you to check if my reasoning was sound. Since $\displaystyle aa^{\dagger}{\psi}_{0} = a\sqrt{1}{\psi}_{1} = {\psi}_{0}$ , we can say $\displaystyle <0|aa^{\dagger}|0> = 1$ And, $\displaystyle <0|aa^{\dagger} \left(a^{\dagger}a + 1 \right) |0>$ $\displaystyle = <0|a \left(a^{\dagger}\right)^2a |0> + <0|aa^{\dagger}|0>$ First term equals 0 (?), second equals 1. So when you substitute all these back into the original equation, you get : 1 + 2(1) = 3 as the answer?
Apr 17th 2008, 10:33 AM   #5

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,494
 Originally Posted by sathys Thanks Dan. I understand most of it, but have a couple of follow-up questions... I'm not completely sure why : $\displaystyle <0|a^{\dagger} = 0$ , is true. My main problem is I'm not very comfortable with manipulating things in Dirac notation (i.e. what you're 'allowed' to do). Do you have a short proof for this? Again, do you have a proof for why the squares are 0, when bracketed by the same states?
Recall that a acting on |0> results in a 0. Thus
$\displaystyle <0|a^{\dagger} = (a|0>)^* = 0$

Recall that $\displaystyle a|n> \propto |n - 1>$ and $\displaystyle a^{\dagger}|n> \propto |n + 1>$.

So
$\displaystyle <0|a^2|0> = <0|0|0> = 0$

and
$\displaystyle <0|\left ( a^{\dagger} \right )^2 |0> = <0|a^{\dagger}k_1|1> = k_1<0|a^{\dagger}|1>$

$\displaystyle = k_1<0|k_2|2> = k_1k_2<0|2>$
where k1 and k2 are the required constants of proportionality. (You can look them up, I'm not going to bother to write them.)

But the spectrum of states are orthogonal, thus <0|2> = 0.

So
$\displaystyle <0|\left ( a^{\dagger} \right )^2 |0> = 0$

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

Apr 17th 2008, 10:38 AM   #6

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,494
 Originally Posted by sathys Now, I'm able to follow your working for the rest of the question, and can vaguely follow the logic to the rest of the question but I would like to run things over with you, just to make sure my reasoning is correct. The next steps are: $\displaystyle <0|a \left (a + a^{\dagger} \right)^2a^{\dagger}|0>$ $\displaystyle = <0|a \left (aa^{\dagger} + a^{\dagger}a \right) a^{\dagger} |0>$ Using the commutator relation $\displaystyle [a,a^{\dagger}]=1$, we get $\displaystyle = <0|a \left (1 + 2a^{\dagger}a \right) a^{\dagger} |0>$ $\displaystyle = <0|aa^{\dagger}|0> + 2<0|aa^{\dagger} \left(a^{\dagger}a + 1 \right) |0>$
Why did you use the commutator again?
$\displaystyle <0|a \left (a + a^{\dagger} \right)^2a^{\dagger}|0>$

$\displaystyle = <0|a \left (1 + 2a^{\dagger}a \right) a^{\dagger} |0>$

So far so good....
$\displaystyle = <0|a a^{\dagger} |0> + 2 <0| aa^{\dagger}aa^{\dagger} |0>$

$\displaystyle = 1 + 2 \cdot 1 = 3$

(I am assuming this answer is correct. I have not checked it explicitly.)

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

Last edited by topsquark; Apr 17th 2008 at 12:03 PM.

Apr 17th 2008, 10:49 AM   #7
Junior Member

Join Date: Apr 2008
Posts: 4
 Originally Posted by topsquark Why did you use the commutator again? $\displaystyle <0|a \left (a + a^{\dagger} \right)^2a^{\dagger}|0>$ $\displaystyle = <0|a \left (1 + 2a^{\dagger}a \right) a^{\dagger} |0>$ So far so good.... $\displaystyle = <0|a a^{\dagger} |0> + 2 <0| aa^{\dagger}a^{\dagger} |0>$ $\displaystyle = 1 + 2 \cdot 1 = 3$ (I am assuming this answer is correct. I have not checked it explicitly.) -Dan
I used the fact that $\displaystyle [a,a^{\dagger}] = aa^{\dagger} - a^{\dagger}a = 1$

Which implies
$\displaystyle aa^{\dagger} = 1 + a^{\dagger}a$

Then sub this back into $\displaystyle aa^{\dagger} + a^{\dagger}a$

$\displaystyle = 1 + 2a^{\dagger}a$

Is this unnecessary work?

Apr 17th 2008, 12:05 PM   #8

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,494
 Originally Posted by sathys I used the fact that $\displaystyle [a,a^{\dagger}] = aa^{\dagger} - a^{\dagger}a = 1$ Which implies $\displaystyle aa^{\dagger} = 1 + a^{\dagger}a$ Then sub this back into $\displaystyle aa^{\dagger} + a^{\dagger}a$ $\displaystyle = 1 + 2a^{\dagger}a$ Is this unnecessary work?
You applied the commutator twice, which usually "undoes" what you did in the first place. You did nothing wrong, you simply didn't do it efficiently. (I recognize that kind of inefficiency very well because I am often guilty of it!)

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Apr 17th 2008, 12:26 PM #9 Junior Member   Join Date: Apr 2008 Posts: 4 Thanks very much Dan...it all makes sense now.

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post toyota11 Kinematics and Dynamics 1 Mar 15th 2012 12:54 PM VincentP Special and General Relativity 0 Nov 23rd 2011 05:27 AM aaaa202 Quantum Physics 1 Jan 4th 2011 09:04 PM CorruptioN Quantum Physics 1 Sep 26th 2010 06:13 AM fasa Advanced Waves and Sound 1 Jan 19th 2009 10:44 AM