Physics Help Forum Ladder operators/Dirac notation

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 Apr 17th 2008, 02:13 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, 04:52 AM   #2

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 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
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Apr 17th 2008, 04:57 AM   #3

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 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
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 Apr 17th 2008, 08:46 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, 09:33 AM   #5

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 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
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Apr 17th 2008, 09:38 AM   #6

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 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
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Last edited by topsquark; Apr 17th 2008 at 11:03 AM.

Apr 17th 2008, 09:49 AM   #7
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 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, 11:05 AM   #8

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 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
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 Apr 17th 2008, 11:26 AM #9 Junior Member   Join Date: Apr 2008 Posts: 4 Thanks very much Dan...it all makes sense now.

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