# 1-D Simple Harmonic Oscillator with Dirac Notation [closed]

I'm having trouble understanding some of the subtleties of working with bras and kets when considering the standard 1-D quantum oscillator.

Say I am given a state vector at $t=0$, $$|\Psi(t=0)\rangle = A \sum_{q=0}^{Q_o} \frac{1}{q+i}|\phi_q\rangle$$

Where $Q_o$ is a positive finite integer.

I am given the number operator $$\hat{N}|\phi_q\rangle = q|\phi_q\rangle$$ where $\hat{N} = \hat{a}^\dagger\hat{a}$, and the energy eigenvalues of the quantum oscillator are known as $$E_q = (q+\frac{1}{2})\hbar\omega_o$$ for $q = 0,1,2,3....$

Now, I am tasked to answer the following questions regarding the state vector $|\Psi(t=0)\rangle$:

[1]

Find an equation for $A$ which normalizes the state vector $|\Psi(t=0)\rangle$

In wave mechanics, the coefficient for normalization is typically found by enforcing that $\int|\Psi|^2 dx = 1$ for all space, and then solving for the coefficient within $\Psi$.

If I am to let $Q_o = 100$, $$|\Psi(t=0)\rangle = |\Psi(0) \rangle = A \sum_{q=0}^{100} \frac{1}{q+i}|\phi_q\rangle$$

I know that a normalized state vector will follow $$\langle \Psi(t)|\Psi(t)\rangle = 1$$

So if I have $$\langle \Psi(0)| = (|\Psi(0)\rangle)^* = A^*\sum_{q=0}^{100} \frac{1}{q-i}\langle\phi_q|$$

then, $$\langle \Psi(0)|\Psi(0)\rangle = (A^*\sum_{q=0}^{100} \frac{1}{q-i}\langle\phi_q|)(A \sum_{q=0}^{100} \frac{1}{q+i}|\phi_q\rangle)=1$$

Now if $A = A^*$, am I able to state that $$A^2\sum_{q=0}^{100}\sum_{q=0}^{100}\frac{1}{q+i}\frac{1}{q-i}\langle\phi_q|\phi_q\rangle = 1$$

where $$\langle\phi_q|\phi_q\rangle = \langle q|\hat{N} |q\rangle = q\langle q|q\rangle = q$$ (as $\langle q|q\rangle = 1$ due to orthonormality), and then solve for $A$?

Or am I horribly mistaken in terms of how bras and kets work?

From what I have so far, I would obtain that $$A^2\sum_{q=0}^{100}\sum_{q=0}^{100}\frac{q}{(q+i)(q-i)} = 1$$ and $$A = \sqrt{\sum_{q=0}^{100}\sum_{q=0}^{100}\frac{(q+i)(q-i)}{q}}$$

Hopefully someone will be able to clarify my misunderstandings here..

[2]

Evaluate $\langle \phi_q|\hat{N}$

Am I able to simply state here that $$\langle \phi_q|\hat{N} = \sum_{q}N_{p,q}\langle \phi_q| = \sum_{q}N_{p,q}\langle q|\hat{a}^\dagger$$

where $N_{p,q}$ are the matrix elements of the number operator $\hat{N}$?

I don't quite know what I am supposed to end up with as an answer or proceed further to work with the sum produced, provided it is even correct.

[3]

What is the state vector for times $t \geq 0$, and the expectation value for $\hat{a}^\dagger$ for these times?

From my understanding, I can simply tack on time dependence to the original state vector. For example, $$|\Psi(t)\rangle = |\phi_q\rangle e^{\frac{-iE_qt}{\hbar}}$$

then, will the expectation value $\hat{a}^\dagger$ just be $$\langle \Psi(t)|\hat{a}^\dagger|\Psi(t)\rangle$$

I'm not sure how to actually calculate this expectation value, but looking at my notes, I see that $$\langle p|\hat{a}^\dagger|q\rangle = \sqrt{q+1}\delta_{p,q+1}$$

Does this imply that I can write $$\langle \hat{a}^\dagger\rangle = \langle \phi_q e^{-i\omega_o t}|\hat{a}^\dagger|\phi_q e^{-i\omega_o t}\rangle = \sqrt{q+1}e^{-i\omega_o t}$$?

Sorry for the length of the question, but hopefully it will make it easy to figure out where I'm going wrong in my approach of these problems. Thanks.

## closed as off-topic by Javier, Gert, Jon Custer, Kyle Kanos, heatherNov 23 '16 at 12:17

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• I'm sorry to say that you shouldn't ask check-my-work type questions here. Very brief comment: the idea for 1) is right, but use different indices for your sums, and you made a mistake doing the inverse of a sum when solving for $A$; also, $|\phi_q\rangle$ is probably the same as $|q\rangle$, so it's already normalized. For 2), take the adjoint of $N|\phi_q\rangle$, it's a pretty trivial answer. For 3), each eigenstate evolves exponentially, but your state is a sum of many eigenstates with different energies. For the expectation value use that $a^\dagger|q\rangle = \sqrt{q+1}|q+1\rangle$. – Javier Nov 22 '16 at 23:32
• Sorry if it is against the rules to post this, but thank you for the response. Will delete the question if necessary. I just see all the time people who post questions with no effort put towards finding their own solution, and how they get torn apart. It is just my way of efficiently learning from my mistakes and correct my way of thinking by showing my steps and thought process. With that said, can you elaborate on what you mean by "inverse of a sum"? I'm not sure what that entails. And when you say $|\phi_q \rangle = |q\rangle$, do you mean that $|\phi_q \rangle$ is normalized already? – bleuofblue Nov 22 '16 at 23:42
• You don't need to delete it; if enough people vote to close it'll be closed. It's great that you show your work, but questions should be about a physics concept of broader interest instead of just an exercise, and I'd say yours is pretty borderline. – Javier Nov 23 '16 at 1:01
• Regarding the inverse of the sum: you have a double sum which schematically is like $\sum a/b$: a sum of fractions. its inverse is $1/\left(\sum a/b\right)$, not $\sum b/a$. And I can't really tell without seeing the definition, but I'd guess that $|\phi_q\rangle$ are the normalized eigenvectors, it's pretty standard notation. – Javier Nov 23 '16 at 1:02
• Ok, so to put things into perspective..... If the $|\phi_q \rangle$ (or $|q\rangle$) are normalized already, does this change the expression in my equation for $A$ to $\langle\phi_q|\phi_q\rangle = 1$ as opposed to $q$? Or am I getting off base with that? – bleuofblue Nov 23 '16 at 2:28

Number 1. In general terms.

A simple harmonic operator is in a state (at t = 0), with state vector

$\Psi (x, 0) = A \sum_{n=0}^{\infty} c^n|\phi_n (x)\rangle$

The normalisation constant A, up to a constant phase, is:

$$A^{-2} = \sum_{n=0}^{\infty} |c|^{2n} =\frac {1}{1-|c|^2}$$

Leads to $$A = \sqrt {1-|c|^2}$$

The time evolved state vector is

$$\Psi (x, t) = A e^{-i\omega t/2} \sum_{n=0}^{\infty} c^ne^{-in\omega t}|\phi_n (x)\rangle$$

Number 3.

{\displaystyle {\begin{aligned}{\hat {x}}&={\sqrt {{\frac {\hbar }{2}}{\frac {1}{m\omega }}}}(a^{\dagger }+a)\\{\hat {p}}&=i{\sqrt {{\frac {\hbar }{2}}m\omega }}(a^{\dagger }-a)~.\end{aligned}}}

This means that a acts on |n⟩ to produce, up to a multiplicative constant, |n–1⟩, and a† acts on |n⟩ to produce |n+1⟩. For this reason, a is called a "lowering operator", and a† a "raising operator". The two operators together are called ladder operators. In quantum field theory, a and a† are alternatively called "annihilation" and "creation" operators because they destroy and create particles, which correspond to our quanta of energy.

Given any energy eigenstate, we can act on it with the lowering operator, a, to produce another eigenstate with ħω less energy. By repeated application of the lowering operator, it seems that we can produce energy eigenstates down to E = −∞. However, since

$${\displaystyle n=\langle n\mid N\mid n\rangle =\langle n\mid a^{\dagger }a\mid n\rangle =\left(a\mid n\rangle \right)^{\dagger }a\mid n\rangle \geqslant 0}$$

• Ok, so here can I make the connection that $c^n = \frac{1}{n+i}$? Further, is a wave function equivalent in every way to a state vector? – bleuofblue Nov 22 '16 at 15:33
• Will see what progress I make tonight in this regard. Thank you for all your time by the way. Another clarification though, to be general about the $c^n$ term from before, when you state that $c^{100} = (\frac{1}{n+i})^{100}$, does this really mean that I would put $c^n = (\frac{1}{n+i})^n$? Sorry to be pedantic here, I just want to ensure I don't misunderstand. But as the state vector includes a sum over $n$, I will need to include each $c$ for each $n$, yes? And $c = \frac{1}{n+i}$ – bleuofblue Nov 22 '16 at 22:41
• Definitely states that $Q_o$ is not infinite. Actually, in the question I was given, I believe the value is 1000. However, maybe I can utilize mathematica to help me with the sum, provided I need to actually compute it vs. just obtain an expression. I will confer with my professor tomorrow about his expectations of this question and report back! Sorry I am going so in depth here, but one more thing I'd like to clear up is how you got to the expression of $A^{-2} = \sum_{0}^{\infty}|c^{2n}|$. It's not immediately obvious to me when I consider the steps I took to try and find $A$. – bleuofblue Nov 22 '16 at 23:07