I have got this problem where I have been given the following wave function: $$\Psi = 0\quad\text{if}~|x| > a\quad\text{and}\quad A(a^2-x^2)\quad \text{if} \quad |x|< a$$

Now the first question is "Determine the time-dependent wave-function. Your answer should be expressed as an infinite sum." How do i express it as an infinite sum, can I just not multiply by $e^{-iEt/(\hbar)}$?

The second question is "What is the probability of finding the particle with energy $E_n = \hbar\omega(n + 1/2)$ at the time $t =2\pi/\omega$?" I don't really get where I shall start, if I express the time dependent wave function as $\psi (x,0)e^{-iEt/\hbar}$ I can substitute for $E$ and $t$, but then how do I find the probability, don't you need an interval to do that?

EDIT: November 6th 2013

I think I have figured out what is meant to do now, this is how I have interrupted the problem with the help I have been given. Let's say for simplicity we want to find the probability of finding the particle with energy $E_1$ at time $t$. Firstly we decompse the wave function into a sum of eigen-functions, then we get that $$\Psi(x,0)=\sum_i^{\infty}c_i\psi_i(x)$$ then we find the time-dependent wave fucntion by doing the following: $$\Psi(x,t)=\sum_i^\infty c_i\psi_i(x)\cdot e^{-iE_nt/\hbar}$$. Since we know that $\vert c_i \vert^2 $ is the proability and $c_i$ is given by: $c_i=\int_{allspace} \psi(x)\psi_i(x)dx $ Now we are intrested in finding $c_1$ and thus we need to know $\psi_1(x)$ which by the formula given by WetSavannaAnimal aka Rod Vance gives the following: $$\psi_1(x)=\frac{1}{\sqrt2}(\frac{m\omega}{\pi\hbar})^\frac{1}{4}\cdot e^{-\frac{m\omega x^2}{2\hbar}}\sqrt{\frac{m\omega}{\hbar}}x^2$$ Substituting this into the formula of $c_i$ for the given wave fucntion and for space being between $-a$ to $a$ and pulling out the constants of the integral yields: $$c_1=A\sqrt{\frac{m\omega}{2\hbar}}(\frac{m\omega}{\pi\hbar})^\frac{1}{4}\int_{-a}^a(a^2-x^2)x^2e^{-\frac{m\omega x^2-2iE_1t}{2\hbar}}$$ after this we just find $\vert c_1 \vert ^2$ and if we want to find the probability for E_n we do the same just a bit more general with $\psi_i(x)$ but the priniple is the same. Is this the right way or at least one way of solving the problem or am I on a total wrong track?


closed as off-topic by Emilio Pisanty, tpg2114, Qmechanic Nov 4 '13 at 7:03

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    $\begingroup$ A time-dependent wavefunction has the form "(psi at time zero) times (e to the -iEt)" only if "psi at time zero" is an energy eigenstate, i.e. predicts that energy value with 100% probability. Do you know about eigenstates yet? $\endgroup$ – Mitchell Porter Nov 3 '13 at 23:17
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    $\begingroup$ Why is this getting close votes? It's a hw question, but it shows enough effort to justify it's existence, in my opinion. $\endgroup$ – Abhimanyu Pallavi Sudhir Nov 4 '13 at 3:52
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    $\begingroup$ @DIMension10 Read the first paragraph (the tldr) of the homework policy. $\endgroup$ – Manishearth Nov 4 '13 at 15:37
  • $\begingroup$ You're on the right track but you should note that your formula for $\psi_1$ is incorrect. $\endgroup$ – Emilio Pisanty Nov 6 '13 at 20:55
  • $\begingroup$ How come? I used what WetSavannaAnimal aka Rod Vance gave me and I checked that $H_1=x$ then I just simplified the expression (the Hermite polynomial) $\endgroup$ – Sindbad Nov 6 '13 at 21:58

I'll help you answer your second question.

But first, there are some difficulties with the problem you've been given. Firstly your question isn't fully specified: you need to have the Hamiltonian itself so you need at least the potential $V(x)$. The expression for $E_n$ you are given bespeaks either a quantum harmonic oscillator or an infinite well. Secondly, the there is a typo in your beginning quantum state. The two inequalities in the definition should read $|x|<a$ and $|x|>a$ (not $|x|<0$, which no $x\in\mathbb{R}$ fulfills!).

So I'll presume an infinite well as this ties in the with Fourier series method of akhmeteli's answer.

As in akmetieli's answer, you expand the quantum state in energy eigenstates $\mathcal{N}^{-1} \cos\left(\left(n + \frac{1}{2}\right)\frac{\pi}{a}x\right);\,n=0,1,2,\cdots$ (when $|x|<a$, nought outside the interval) corresoponding to the energies $E_n$ (here $\mathcal{N}$ is the normalization to make $\int_{-a}^a|\psi|^2dx=1$.

Note that this is a discrete series. The energy doesn't take on a continuous values, so the magnitude squared of your Fourier series (not integral as for continuously varying energy) weights are the probabilities, not probability densities, that the quantum state will be found in each energy.

So when you do your Fourier series, you should check that $\sum_n |w_n|^2 = 1$, where $w_n$ are the Fourier series weights.

Now the probability $|w_n|^2$ does not vary with time. The phase of $w_n$ does, so the energy eigenstates interfere with one another to give a time varying wavefunction, but the probabilities to be in each energy eigenstate are constant. So you don't even need to know the time when you calculate your probability.

This should let you finish your question.

Next stage: Since you are dealing with the quantum harmonic oscillator, your energy eigenstates are the following set:

$$\psi_n(x) = \frac{1}{\sqrt{2^n\,n!}} \cdot \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \cdot e^{ - \frac{m\omega x^2}{2 \hbar}} \cdot H_n\left(\sqrt{\frac{m\omega}{\hbar}} x \right), \qquad n = 0,1,2,\ldots.\qquad(1) $$

where $H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}\left(e^{-x^2}\right)$ is the $n^{th}$ Hermite polynomial.

These eigenfunctions are orthogonal in the sense that:

$$\left<\psi_n, \psi_m\right> \stackrel{def}{=} \int_{-\infty}^\infty \psi_n(x)^* \psi_m(x) {\rm d}x = \delta_{m,n}\qquad (2)$$

i.e. the "inner product" is nought for different discrete energy eigenfunctions and 1 for $n=m$, so each eigenstate is "normalized" or said to have unit "length" in the Hilbert space spanned by the energy eigenstates (don't worry if you don't understand all of this; these are the kinds of statements that'll become more and more wonted to you if you keep at your self study). The above orthogonality is the key to the kind of decomposition that Akhmeteli and I have spoken of: any initial quantum state $\psi(x)$ can be resolved into its energy eigenstates by assuming:

$$\psi(x) = \sum\limits_{m=0}^\infty w_m \,\psi_m(x)\qquad(3)$$

then multiplying both sides of (3) by the $m^{th}$ eigenstate in turn and integrating over the whole real interval, applying (2) to find (noting that we can integrate the series termwise):

$$w_m = \int_{-\infty}^\infty \psi(x)\, \psi(m)\, {\rm d}x \,\qquad(4)$$

Now it is the field of spectral theory that shows that our set of eigenfunctions (1) is complete, i.e. that a sum of the kind (3) can indeed represent (in the appropriate measure theoretic sense) any piecewise continuous $\psi(x)$ fulfilling $\int_{-\infty}^\infty |\psi(x)|^2 {\rm d}x < \infty$, which of course is true for valid quantum states since we must have $\int_{-\infty}^\infty |\psi(x)|^2 {\rm d}x =1$ (this is just a necessary condition for $|\psi(x)|^2$ to be a probability density in $x$).

You will ultimately learn that this "orthogonality" is a property of all eigenfunctions of any quantum observable, not only the Hamiltonian. This result comes to us from Sturm Liouville theory and is owing to the self-adjointhood of quantum observables (more generally it holds for any normal operator - one that commutes with its own adjoint).

Lastly, note that, since $\psi_n(x)$ is the $n^{th}$ energy eigenfunction, its full space and time variation is $\psi_n(x, t) = \psi_n(x) \exp(-i E_n t/\hbar)$. So once you've resolved your beginning quantum state into a superposition like (3), you can write down its general time dependence:

$$\psi(x, t) = \sum\limits_{m=0}^\infty \left(w_m \,\psi_m(x)\,e^{-i \frac{E_n}{\hbar} t}\right)\qquad(5)$$

This should let you get a bit further!

  • $\begingroup$ Is there another way around the problem other than using the Fourier series (since I am not familiar with that yet)? Plus we have been given the Hamoltionian as: H=p^2/2m + 1/2*m\omega x^2, where p and x are operators. $\endgroup$ – Sindbad Nov 3 '13 at 22:51
  • $\begingroup$ @Axcelneo then this is a quantum harmonic oscillator. Therefore, you're not going to be using FOurier series. You're going to expand into a general orthogonal series of the Hermite-Gauss modes as in the first section of en.wikipedia.org/wiki/Quantum_harmonic_oscillator. They are all orthogonal, so calculating the weights is done likewise with Fourier series. Are you sure you haven't been shown the decomposition in class? Have a look in your text - which one are you using? $\endgroup$ – WetSavannaAnimal Nov 3 '13 at 23:16
  • $\begingroup$ I am self learning the subject so I have not been shown anything. I use lectures on the internet and The Feynman Lectures on Physics Volume III: Quantum Mechanics. $\endgroup$ – Sindbad Nov 3 '13 at 23:21
  • $\begingroup$ @Axcelneo Good for you. In that case I'll find a reference for you. Feynman is excellent, but he doesn't go through the QHO IIRC. Can you get to a library - find Merzebacher or Griffiths? $\endgroup$ – WetSavannaAnimal Nov 3 '13 at 23:23
  • $\begingroup$ Sure, but I like to get a feel for the subject before I start going in depth with the littrature, that is, I like to be able to calculate simple problems like the one above before picking up a book on the subject. Feynman just happend to be around. $\endgroup$ – Sindbad Nov 3 '13 at 23:27

You cannot "just multiply" by the exponent as the function is not an eigenstate of the free Hamiltonian (I guess the free Hamiltonian (no potential) is assumed in the problem). So you should expand the initial function into a Fourier series (a sum of eigenstates of momentum) and only then multiply the terms in the expansion by the exponents with the relevant energies.

EDIT (11/3/2013): As the Hamiltonian turns out to be that of an oscillator, you should expand the function in a series in the eigenstates of the Hamiltonian. Please see http://www.gauge-institute.org/delta/HermiteDelta.pdf , formulas at the end of p.10 and at the beginning of page 11.

  • $\begingroup$ Alright, but exactly how do I do that? $\endgroup$ – Sindbad Nov 3 '13 at 22:41
  • $\begingroup$ @Axcelneo: First of all, according to your comment on WetSavanna's answer, my guess about the Hamiltonian being free turned out to be wrong, so you should expand the wavefunction into a series in eigenstates of the Hamiltonian. I cannot describe the procedure here. I guess your professor or your book offered some hints. $\endgroup$ – akhmeteli Nov 3 '13 at 23:28

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