Energy of perturbed harmonic oscillator

  • #1
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Homework Statement


For a quantum harmonic oscillator in an electric field, using ##\hat{V}=q\epsilon\hat{x}##, with the following trial state: $$|\psi\rangle=|0\rangle+b|1\rangle$$
Show that the energy can be written as $$E=\frac{\frac{\hbar \omega}{2}+\frac{3b^2\hbar\omega}{2}+\frac{\sqrt{2}bq\epsilon}{\alpha}}{1+b^2}$$

Homework Equations



##\hat{x}=\frac{1}{\sqrt{2}\alpha}(a^{\dagger}+a)##

The Attempt at a Solution



There are two things that are confusing me:

(1) how to distribute the brakets in the following:

##\langle \psi | ~~= \langle 0 | +b^*\langle 1 |##
##E^{(1)}_{\psi}=\frac{q\epsilon}{\sqrt{2}\alpha}\langle \psi|a^{\dagger}+a|\psi\rangle##
##~~~~~~~\stackrel{?}{=}\frac{q\epsilon}{\sqrt{2}\alpha}[(\langle 0|+b^*\langle 1|)a^{\dagger}(|0\rangle+b|1\rangle)+(\langle 0|+b^*\langle 1|)a(|0\rangle+b|1\rangle)]##
##~~~~~~~=\frac{q\epsilon}{\sqrt{2}\alpha}[\langle 0|1 \rangle+\sqrt{2}b\langle 0|2 \rangle+b^*\langle 1|1 \rangle+\sqrt{2}b^*b\langle 1|2 \rangle+b\langle 0|0 \rangle+b^*b\langle 1 |0 \rangle]##
##~~~~~~~=\frac{q\epsilon(b^*+b)}{\sqrt{2}\alpha}##

Is this how you would distribute it? I get the feeling it is completely wrong..

(2) What is the unperturbed energy of the state ##|\psi \rangle##

Would it be ##E_{\psi}^{(0)}=\frac{1}{2}\hbar\omega+\frac{3}{2}b\hbar\omega## ?

Any guidance is appreciated!
 

Answers and Replies

  • #2
blue_leaf77
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Show that the energy can be written as $$E=\frac{\frac{\hbar \omega}{2}+\frac{3b^2\hbar\omega}{2}+\frac{\sqrt{2}bq\epsilon}{\alpha}}{1+b^2}$$
First of all, are you sure you copied the above equation correctly, as it says in whatever source you took from? Because if ##b## is complex, as it generally is, that equation cannot seem to be true because it will yield a complex value.
 
  • #3
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Hi, yes I am sure I have copied it correctly, here is a snapshot of the question. Please note I have swapped the a with b, to avoid any confusion.

Physics forum perturbation Qn.PNG


Maybe it is assumed that a is real?
 
  • #4
blue_leaf77
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Yeah, it seems like the question maker forgot to state that ##a## is real, otherwise the minimization problem cannot be carried out with only one equation because a complex number consists of two real numbers.
Anyway, I don't see anything wrong with
(1) how to distribute the brakets in the following:

##\langle \psi | ~~= \langle 0 | +b^*\langle 1 |##
##E^{(1)}_{\psi}=\frac{q\epsilon}{\sqrt{2}\alpha}\langle \psi|a^{\dagger}+a|\psi\rangle##
##~~~~~~~\stackrel{?}{=}\frac{q\epsilon}{\sqrt{2}\alpha}[(\langle 0|+b^*\langle 1|)a^{\dagger}(|0\rangle+b|1\rangle)+(\langle 0|+b^*\langle 1|)a(|0\rangle+b|1\rangle)]##
##~~~~~~~=\frac{q\epsilon}{\sqrt{2}\alpha}[\langle 0|1 \rangle+\sqrt{2}b\langle 0|2 \rangle+b^*\langle 1|1 \rangle+\sqrt{2}b^*b\langle 1|2 \rangle+b\langle 0|0 \rangle+b^*b\langle 1 |0 \rangle]##
##~~~~~~~=\frac{q\epsilon(b^*+b)}{\sqrt{2}\alpha}##
The only small mistake lies in your calculation of the unperturbed "energy". Note that the trial function is a superposition state, therefore you cannot associate any energy eigenvalue to it. What you can do is to calculate the expectation value of energy. This is the reason why you got only ##b## in the 2nd term of ##E_{\psi}^{(0)}## instead of ##b^2## as it should be. Once you correct this and take normalization into account, you should be able to prove what is asked.
 
Last edited:
  • #5
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Thanks for the help blue leaf! I was able to prove the statement.

I have another question, what does ##| 0\rangle## represent? is it the column vector ##(0,1,0,0,\dots)^T##? And what do these vectors actually mean? If you solve the QHO using integrals and all that ugly stuff, you find that the eigenfunctions are in the form of hermite polynomials, how are these and ##| i \rangle## related ?

If you prefer, I can start a new thread with this question.
 
  • #6
blue_leaf77
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State in quantum mechanics is mathematically represented by a vector in a vector space. This vector is commonly denoted as a braket ##|a\rangle## in QM. From linear algebra, an arbitrary state can be represented as a linear combination of basis states provided these basis states have the same dimensionality as the state being represented does. Now there are many vectors spaces whose basis states can be used to represent any arbitrary state of the same dimensionality as these vector spaces, one of them is the continuous position space spanned by the basis states ##|x\rangle##. If you represent an eigenstate ##|m\rangle## of a QHO in terms of ##|x\rangle##'s, the expansion coefficient ##\langle x|m \rangle## is the one which you will find to exhibit those Hermite polynomial thing.
 

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