Useful Raising/Lowering Operator Equation, Quick derivation help

In summary, the formula states that the sum of the squares of the derivatives of two operators is the same as the derivative of the third operator. If two operators are equal, then their derivatives are also equal.
  • #1
logic smogic
56
0
Problem

Show that,

[tex]a (a^{ \dagger})^{n} = n (a^{\dagger})^{n-1}+(a^{\dagger})^{n} a[/tex]

Formulae

[tex] a = \sqrt{\frac{m \omega}{2 \hbar}}(x+\frac{\imath p}{m \omega})[/tex]

[tex] a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}}(x-\frac{\imath p}{m \omega})[/tex]

[tex] [a,a^{\dagger}]= a a^{\dagger}-a^{\dagger}a=1[/tex]

Attempt

This is just one step in a long derivation from another problem. The author (Goswami, pg 147) uses it without proof, but I would like to modify it – and hence I need to understand where he got it from.

I can see how you might pull [tex] a^{\dagger}[/tex]’s out from the first and third term, and try to use the commutation relation, but the n’s don’t seem to work out right.

Any thoughts?
 
Last edited:
Physics news on Phys.org
  • #2
The formula you are trying to prove is the same as [a,(a^+)^n]=n*(a^+)^(n-1), right? For any three operators A,B and C we know [A,BC]=[A,B]C+B[A,C]. This looks like the product rule for differentiation with [A, acting like a derivative. You can extend this to any number of operators, [A,BCDEF...]. So what does this say about [A,B^n] when [A,B]=1? BTW I think [a,a^+]=1, not the order you wrote.
 
  • #3
Ah, I didn't think to look up some commutator identities. And, yes, we proved the differentiation/commutation relationship last homework assignment. Right, I'll try it again. Thanks.
 
  • #4
So then,

[tex][a^{2}, (a^{\dagger})^{n}] = a [a, (a^{\dagger})^{n}]+[a, (a^{\dagger})^{n}]a= a n (a^{\dagger})^{n-1} + n (a^{\dagger})^{n-1} a[/tex]

Consider,

[tex]|n>=\frac{1}{\sqrt{n!}}(a^{\dagger})^{n}|0>[/tex]

where

[tex]a|0>=0[/tex]

then,

[tex]a^{2}|n>=\frac{1}{\sqrt{n!}}a^{2}( a^{\dagger})^{n}|0> [/tex]
[tex]= \frac{1}{\sqrt{n!}}( a n (a^{\dagger})^{n-1} + n (a^{\dagger})^{n-1} a
)| 0>[/tex]
[tex]= \frac{1}{\sqrt{n!}} a n (a^{\dagger})^{n-1}| 0>[/tex]
[tex]=\frac{1}{\sqrt{n!}} n (n-1) (a^{\dagger})^{n-2}|0>[/tex]
[tex]= \sqrt {n (n-1)} |n-2>[/tex]

I know that I may have not provided enough of a primer in the problem to have anyone check this, but just puttin’ it out there!
 
Last edited:

1. What is a raising/lowering operator equation?

A raising/lowering operator equation is a mathematical relationship between two operators that act on a particular mathematical space. These equations are commonly used in quantum mechanics to describe the behavior of particles and their energy levels.

2. How do you use a raising/lowering operator equation?

To use a raising/lowering operator equation, you first need to identify the operators involved and the mathematical space they act on. Then, you can apply the equation to a given state vector in that space to obtain a new state vector with modified properties.

3. What is the purpose of a raising/lowering operator equation?

The purpose of a raising/lowering operator equation is to describe the behavior of particles in quantum mechanics, specifically their energy levels. These equations help to determine the possible states and transitions of a particle, providing valuable information for understanding its behavior.

4. How is a raising/lowering operator equation derived?

A raising/lowering operator equation is derived using mathematical techniques such as linear algebra and differential equations. These equations are based on the properties of the operators involved and their relationships to each other.

5. Can a raising/lowering operator equation be used in other fields besides quantum mechanics?

While raising/lowering operator equations are most commonly used in quantum mechanics, they can also be applied in other fields such as statistical mechanics and electromagnetism. These equations can be used to describe the behavior of particles and fields in these areas of physics as well.

Similar threads

Replies
27
Views
2K
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
1K
Replies
11
Views
1K
  • Advanced Physics Homework Help
Replies
24
Views
679
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
26
Views
2K
Replies
2
Views
3K
Back
Top