Ladder operators and matrix elements...

In summary, the conversation is about using ladder operators to simplify the calculation of matrix elements, specifically calculating <u+2|P2|u> and <u+1| X3|u>. The value is 0 for both i and ii when u is different on both sides. The conversation also discusses the use of a and a+ operators, as well as the occupation number operator, [a,a+]=1. The steps to compute expressions such as <u+2|P2|u> are explained, and the only non-zero result is obtained for <u+2|a†a†|u>.
  • #1
Activeuser
8
0
Please I need your help in such problems..
in terms of ladder operators to simplify the calculation of matrix elements... calculate those
i) <u+2|P2|u>
ii) <u+1| X3|u>
If u is different in both sides, then the value is 0? is it right it is 0 fir both i and ii?
when exactly equals 0, please explain slowly, my background is chemistry and not physics..
Thank you.
 
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  • #2
You'll need to supply a bit more context...

I guess P and X are momentum and position operators respectively? But what is your "u"?
 
  • #3
Well, in terms of ladder operators, X= (h/2mw)1/2(a+a+) you can say = constant (a+a+) it is a real
P=i(hwm/2)1/2 (a-a+) or constant (a-a+)
about u and u+1 are functions; some books write them like this and others write PSI or PHI of subscript u and u+1;I couldn't type that way as there is no symbols here.
a+a=n occupation number operator , [a,a+]=1

If needed I can give an example I know..
 
  • #4
Activeuser said:
Well, in terms of ladder operators, X= (h/2mw)1/2(a+a+) you can say = constant (a+a+) it is a real
P=i(hwm/2)1/2 (a-a+) or constant (a-a+)
about u and u+1 are functions; some books write them like this and others write PSI or PHI of subscript u and u+1;I couldn't type that way as there is no symbols here.
a+a=n occupation number operator , [a,a+]=1

If needed I can give an example I know..

I'm assuming that [itex]|u\rangle[/itex] means the state such that the number operator [itex]N = a^\dagger a[/itex] has value [itex]u[/itex]? In that case, it's completely straight-forward, if tedious, to compute expressions such as

[itex]\langle u+2 | P^2 | u \rangle[/itex]

Just write [itex]P[/itex] in terms of [itex]a[/itex] and [itex]a^\dagger[/itex], and apply the rules:

[itex]a |u\rangle = \sqrt{u} |u - 1\rangle[/itex]
[itex]a^\dagger |u\rangle = \sqrt{u+1} |u+1\rangle[/itex]
[itex] \langle n | m \rangle = 0[/itex] if [itex]n \neq m[/itex]
[itex]\langle n | n \rangle = 1[/itex]

So [itex]P^2 = A (a^2 - a a^\dagger - a^\dagger a + (a^\dagger)^2)[/itex] (for some constant [itex]A[/itex])
[itex]P^2 | u \rangle = A (a^2 |u \rangle - a a^\dagger |u\rangle - a^\dagger a |u\rangle + (a^\dagger)^2 |u\rangle)[/itex]

Then you just work out what [itex]a^2 |u\rangle[/itex] is, etc.
 
  • #5
ok great. I know till this step, my question is to separate these operations and get the final out put like this
<u+2|aa|u> - <u+2|aa+|u> - <u+1|a+a|u>... so on
and solve each one on its own, when I get any different matrix output, the term equals 0; like <u+1|u> or <u+2|u-1>... right? Cuz my confusion is about this part.

the other part is how to work a3,, if aaa, the 1st one from left operates with the left , and the 1st a from right operates with the right function of the matrix, the middle a operator works with which function.. as it gives different values..
I am sorry for my silly questions, because this what I miss and can not find in books; it seems like a bases that I do not have.
 
  • #6
Well, just look at the term

[itex]\langle u+2|a\ a|u \rangle[/itex]

Use the rule: [itex]a |u\rangle = \sqrt{u} |u - 1\rangle[/itex]. So that term simplifies(?) to:

[itex]\langle u+2|a\ a|u \rangle = \sqrt{u} \langle u + 2 | a | u-1\rangle[/itex]

Use the same rule again: [itex]a |u-1\rangle = \sqrt{u-1} |u - 2\rangle[/itex]. So we have:

[itex]\langle u+2|a\ a|u \rangle = \sqrt{u(u-1)}| \langle u + 2 | u-2\rangle[/itex]

which is zero.

The only term you're going to get a nonzero result for is:

[itex]\langle u+2|a^\dagger\ a^\dagger|u \rangle[/itex]
 
  • #7
By the way, this should probably be in the Advanced Physics Homework section.
 
  • #8
Thank you so much for this clarification.. it is very helpful.
Next time I will follow the right section.. new here.:smile:
 

Related to Ladder operators and matrix elements...

What are ladder operators in quantum mechanics?

Ladder operators are mathematical operators used in quantum mechanics that allow for the calculation of energy levels and transition probabilities in a quantum system. They are used to describe the behavior of particles in a quantum system and are essential in solving the Schrödinger equation.

What is the significance of matrix elements in quantum mechanics?

Matrix elements in quantum mechanics represent the transition probabilities between different quantum states. They are used to calculate the probability of a particle transitioning from one energy state to another, and play a crucial role in understanding the behavior of quantum systems.

How do ladder operators and matrix elements relate to each other?

Ladder operators and matrix elements are closely related in quantum mechanics. Ladder operators are used to calculate matrix elements, which represent the transition probabilities between different quantum states. They are both fundamental concepts in quantum mechanics and are essential for understanding the behavior of quantum systems.

Can ladder operators and matrix elements be used in other areas of science?

While ladder operators and matrix elements are most commonly used in quantum mechanics, they also have applications in other areas of science such as solid state physics and statistical mechanics. They are powerful mathematical tools for describing the behavior of particles and systems at the quantum level.

Are there any limitations to using ladder operators and matrix elements in quantum mechanics?

While ladder operators and matrix elements are valuable tools in quantum mechanics, they do have limitations. They are only applicable to systems with discrete energy levels, and cannot accurately describe the behavior of particles in continuous energy systems. Additionally, they are only valid in non-relativistic quantum mechanics and cannot be used in relativistic quantum mechanics.

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