Raising and lowering operators homework

In summary, the conversation discusses difficulties with finding the energies of stationary states for a particle with a Hamiltonian expressed as H = A(a+a) + B(aa+), where A and B are constants and a+ and a are the raising and lowering operators. The operators also satisfy the commutation relation [a, a+] = 1. The conversation includes attempts to use these equations to find the energies, with suggestions to group like terms and use the raising operator to find all energies and excited states. The general formula for the energies is En = (n+1)(A+B), with a mention of a possible error in the calculation of the ground state energy.
  • #1
holden
30
0
Having a lot of trouble with this one. I'm given that the Hamiltonian of a certain particle can be expressed by H = A(a+a) + B(aa+), where A and B are constants and a+ and a are the raising and lowering operators, respectively. I'm supposed to find the energies of the stationary states for the particle.

I'm also given that the operators satisfy the communtation relation [a, a+] = 1. So from this I'm getting [tex]aa_+\psi - a_+a\psi = \psi[/tex]. From the Hamiltonian I have [tex]A(a_+a)\psi + B(aa_+)\psi = E\psi[/tex]. I tried using the first equation to replace one of the terms on the left side in the second, to get [tex]A(aa_+\psi - \psi) + B(aa_+\psi) = E\psi[/tex]... but I don't see how this really helps. I'm stuck on what to do next, so any help would be appreciated :)
 
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  • #2
Anybody have a clue on what I could do next? Even an inkling of an idea? Because I'm really lost. Do I need to provide more info? Thanks :)
 
  • #3
holden said:
Anybody have a clue on what I could do next? Even an inkling of an idea? Because I'm really lost. Do I need to provide more info? Thanks :)
Looks like nobody is helping so here are some thoughts

Recall that for the h.o. [itex] H = (a_+a_- +{1 \over2}) = a_- a_+ - { 1\over 2}) [/itex](in units of [itex]\hbar \omega [/itex]).

Then one shows that is [itex] \psi_n [/itex] is a solution with energy E_n, [itex] a_+ \psi_n [/itex] is also a asolution, with energy E_n + 1, and [itex] a_- \psi_n [/itex] is a solution with energy E_n - 1.

Then one imposes that for the ground state, [itex] a_- \psi_0 = 0 [/itex] and one uses [itex] H \psi_0 = E_0 [/itex] to find E_0 and the ground state wavefunction. Then one uses the raising operator to find all the energies and the excited states.

Just go through exactly all the same steps. The difference is that in your case, you will get a slightly different relation for the relation between the energy of psi_n and of [itex] a_+ \psi_n [/itex] and a different expression for the ground state energy and so on. But the steps are all the same.

I hope this helps.

Patrick
 
  • #4
holden said:
I tried using the first equation to replace one of the terms on the left side in the second, to get [tex]A(aa_+\psi - \psi) + B(aa_+\psi) = E\psi[/tex]... but I don't see how this really helps. I'm stuck on what to do next, so any help would be appreciated :)
It would probably help to group like terms. I.E.

[tex](A + B) a a_+ \psi = (E - A) \psi[/tex]

[tex]a a_+ \psi = \frac{E - A}{A+B} \psi[/tex]
 
  • #5
Thanks guys. I think I did it in a convoluted way (before I saw your post, Hurkyl), but I ended up getting that a- operating on a state of energy E would give a state of energy E - A - B, and a+ operating on a state of energy E would give a state of energy E + A + B, which seems to make at least a little sense.

Using this, for the ground state I got E0 = A + B. So that means E1 = 2A + 2B, etc.. so my question is when do I know when to stop? Am I just looking for a general formula, which would just be En = (n+1)(A+B)? Or have I screwed it up.

Thanks so much for the help. This was a fun problem once I knew what I was doing :)
 
  • #6
holden said:
Thanks guys. I think I did it in a convoluted way (before I saw your post, Hurkyl), but I ended up getting that a- operating on a state of energy E would give a state of energy E - A - B, and a+ operating on a state of energy E would give a state of energy E + A + B, which seems to make at least a little sense.

Using this, for the ground state I got E0 = A + B. So that means E1 = 2A + 2B, etc.. so my question is when do I know when to stop? Am I just looking for a general formula, which would just be En = (n+1)(A+B)? Or have I screwed it up.

Thanks so much for the help. This was a fun problem once I knew what I was doing :)
Are you sure your ground state energy is right?
The hamiltonian you gave in your first post was [itex] A a_+ a_- + B a_- a_+ = (A+B) a_+ a_- + B [/itex], right? Applied on the ground state the lowering operator should give zero so that the ground state energy should be B, no?

The general formula will indeed be of the simple form [itex] C_1 + C_2 n [/itex] where C1 and C2 will be functions of A and B.

Patrick
 
  • #7
nrqed said:
The hamiltonian you gave in your first post was [itex] A a_+ a_- + B a_- a_+ = (A+B) a_+ a_- + B [/itex], right?
I'm not sure where the right half of that equation is coming from :(. I got the ground state energy by just saying [tex]a_-\psi_0 = 0[/tex] and [tex][A(a_+a_-) + B(a_-a_+)]\psi_0 = E_0[/tex], and since a- operating on a state of energy E gives a state of energy (E - A - B), I reasoned that a- operating on E0 must give E0 - A - B = 0. So E0 would be A + B. I probably did mess up somewhere though =/
 
  • #8
holden said:
I'm not sure where the right half of that equation is coming from :(.
I just rewrote [itex] a_- a_+ = a_+ a_- + 1 [/itex] in your original hamiltonian.
I got the ground state energy by just saying [tex]a_-\psi_0 = 0[/tex] and [tex][A(a_+a_-) + B(a_-a_+)]\psi_0 = E_0[/tex], and since a- operating on a state of energy E gives a state of energy (E - A - B), I reasoned that a- operating on E0 must give E0 - A - B = 0. So E0 would be A + B. I probably did mess up somewhere though =/
I see what you did. But that's not what we do in the usual harmonic oscillator problem. We assume that [itex] a_- \psi_0 0[/itex] without saying anything about the energy. If we used the same reasoning you used, we would say [itex] E_0 - \hbar \omega =0 [/itex] so the ground state energy would be [itex] \hbar \omega [/itex] but that's not the correct answer. Instead, we assume [itex] a_- \psi_0 = 0 [/itex] without any condition on the energy. We get the energy by applying the hamiltonian.

Hope this makes sense

Patrick
 
  • #9
OK, I see what you're saying. But since I don't know the lowering function explicitly here.. how can I calculate [tex]a_-\psi_0[/tex]?
 
  • #10
holden said:
OK, I see what you're saying. But since I don't know the lowering function explicitly here.. how can I calculate [tex]a_-\psi_0[/tex]?
we simply *define* this to be zero (so that there is a state of minimum energy). If there was not such a state of lower energy, the particle could jump down forever, and we would have a source of infinite energy.

So the key point is that we *define* psi_0 by imposing [itex] a_- \psi_0 = 0 [/itex]. *then* you go back to the hamiltonian which you apply on the ground state and that gives you the ground state energy. If you go back to the harmonic oscillator, this is the logic followed.

Hope this makes sense

Patrick
 
  • #11
OK, maybe we didn't use that approach in my class because I can't find anything about it in my notes or text. So if we apply the hamiltonian to the ground state.. you get [tex][A(a_+a_-) + B(a_-a_+)]\psi_0 = E_0[/tex]. So since [tex]a_-\psi_0 = 0[/tex], you're just left with [tex]Ba_-(a_+\psi_0) = E_0[/tex].. which is [tex]Ba_-(E_1 + A + B) = E_0[/tex] and then [tex]BE_0 = E_0[/tex], which is clearly not right. Where did I go wrong?
 
  • #12
holden said:
OK, maybe we didn't use that approach in my class because I can't find anything about it in my notes or text. So if we apply the hamiltonian to the ground state.. you get [tex][A(a_+a_-) + B(a_-a_+)]\psi_0 = E_0[/tex]. So since [tex]a_-\psi_0 = 0[/tex], you're just left with [tex]Ba_-(a_+\psi_0) = E_0[/tex].. which is [tex]Ba_-(E_1 + A + B) = E_0[/tex] and then [tex]BE_0 = E_0[/tex], which is clearly not right. Where did I go wrong?
:confused: [itex] a_+ \psi_0 [/itex] is *not* [itex] (E_1 +A + B) \psi_0 [/itex]! All you know is that [itex] a_+ \psi_0 [/itex] gives a state [itex ] \psi_1 [/itex] with an energy E_0 + A + B (I did not check this part of your calculation, I assume it's right).

By the way, you must be careful about not dropping the wavefunction (which you did in what you wrote).

No, what you do is to use the commutator to rewrite [itex] a_- a_+ = a_+ a_- +1 [/itex] which *then* you apply to the ground state wavefunction. At this stage of the calculation you certainly don't want to apply the raising operator and get psi_1 all mixed in in your calculation.

What textbook are you using? You could see what I mean by looking at their derivation of the ground state energy.

Patrick
 
  • #13
Sigh. Sorry. I feel like an idiot.

OK. Thanks for spelling that out for me. So I did what you said and got [tex]a_-a_+\psi_0 = \psi_0[/tex]. Applying the Hamiltonian and then using that, you get:

[tex][A(a_+a_-) + B(a_-a_+)]\psi_0 = E_0\psi_0[/tex]
[tex][(A(a_+a_-) + B]\psi_0 = E_0\psi_0[/tex]
[tex]E_0\psi_0 = B\psi_0[/tex]

So E0 is just B. E1 is A + 2B, E2 is 2A + 3B, etc., and in general it's just En = An + B(n+1). I *think* this is what you were saying earlier.
 

Related to Raising and lowering operators homework

1. What are raising and lowering operators and why are they important in quantum mechanics?

Raising and lowering operators are mathematical operations that are used to describe the behavior of quantum mechanical systems. They are important because they allow us to study the energy levels and properties of particles in a quantum system.

2. How do raising and lowering operators work?

Raising and lowering operators are defined as operators that change the energy state of a system by either increasing or decreasing it by a fixed amount. They act on a wave function, and when applied, they either raise or lower the energy level of that wave function.

3. What are the properties of raising and lowering operators?

Raising and lowering operators have several important properties, including the fact that they are hermitian conjugates of each other, they commute with each other, and they obey certain algebraic relationships. These properties are essential in understanding the behavior of quantum systems.

4. Can raising and lowering operators be used in other areas of physics?

Yes, raising and lowering operators have applications in other areas of physics, such as in statistical mechanics, quantum field theory, and even in classical mechanics. They are a powerful tool for describing the dynamics of physical systems and are widely used in various fields of physics.

5. How can I use raising and lowering operators in my homework?

Raising and lowering operators are commonly used in quantum mechanics homework to solve problems related to energy levels and transitions between states. They can also be used to derive equations and solve problems in other areas of physics. It is important to understand the properties and applications of these operators to effectively use them in your homework assignments.

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