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Apologies if this has been asked already

  1. For the 1D particle-in-a-box example, how do we determine the weights of each eigenfunction in the general (time-dependent) solution that fully describes the system? Do we need extra initial conditions for this and if so, what might these conditions be and how might we measure/determine them?

  2. Also, doesn't each eigenstate have constant energy by definition? If so, then shouldn't the momentum of the particle in an eigenstate have a constant magnitude? According to Wikipedia, momentum for the 1D particle-in-a-box has a distribution: https://en.wikipedia.org/wiki/Particle_in_a_box#Position_and_momentum_probability_distributions

  3. Finally, according to QM, $\langle \psi|O|\psi\rangle$ can be used to determine the expectation of observable O. Does QM similarly provide a way to determine the distribution of the observable? If not, it seems the theory is rather incomplete

Thanks in advance

Edit

Thanks for the answers all! A couple edits

  1. Assume psi(0) is unknown, which I'd imagine would generally be the case in an experiment. What would we then expect the general wavefunction to look like? Are all possible coefficient vectors (c1,c2,...,cn) equally likely in practice, or are some combinations more likely than others?

  2. Momentum of particle in a box explains it - QM is incompatible with de Broglie's matter wave theory apparently. Good to know. Also although energy states are quantized, this does not appear to be the case for kinetic energy.

  3. symanzik138 and freecharly answered this one; it makes sense now.

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  • $\begingroup$ I would like to answer your question two in a short manner, that the system you're looking at here is not a free particle system (the domain is only 0 to L), so the energy and momentum relationship is a non trivial one. $\endgroup$ – user35952 Oct 26 '16 at 7:05
  • $\begingroup$ Tip: Let's not have posts look like revision histories. $\endgroup$ – Qmechanic Oct 26 '16 at 8:45
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1) A time dependent state can be written as a superposition of the eigenstates of the Hamiltonian. I am assuming that the Hamiltonian is a constant one; does not change with time. The coefficients or weights of the eigenstates are complex. so, for any state $\psi$,

$\psi(t) = c_1 \phi_1 + c_2 \phi_2 + \dots $

where $\phi_i$ are the normalized eigenstates of the Hamiltonian. The coefficients

$c_i(t) = \langle \phi_i|\psi(t) \rangle$

are time dependent.

And the time dependence has the precise form:

$c_i(t)=\exp(-\imath E_i t/\hbar) c_i(0)$

where $E_i$ is the energy of the state $\phi_i$. Thus given the initial coefficients $c_i(0)$, you can determine the coefficients $c_i(t)$ and therefore the state at any other instant $t$. Equivalently you can take $\psi(0)$ also as the initial condition, as $c_i(0)$ can be inferred from $c_i(0)=\langle \phi_i|\psi(0) \rangle$.

2) Constant energy does not mean a constant momentum. This is because the box is not translation invariant. However, the magnitude of the momentum is a constant (in the case of the simple 1D quantum well. This is not true in general). You can in fact write the energy eigenstate as a linear combination of states of momenta $p$ and $-p$. So momentum distribution is nonzero at $\pm p$ only.

3) Distribution of an observable can be interpreted as follows. The probability of finding a value of $o$ for a measurement (say position or momentum) on a system in $\psi$. Such measurements are associated with a corresponding operator $O$. $p(o) = |\langle\psi | \phi_o\rangle|^2$ if $o$ is an eigenvalue of $O$, and $p(o)=0$ otherwise. $\phi_o$ is the normalized eigenstate of $O$ withe eigenvalue $o$. (I am assume no degeneracy).

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Not an expert by any means, as I am also just beginning to learn quantum mechanics... But I will share my thoughts, and hopefully others can correct me if I am wrong.

(1) To obtain time dependent solutions for the particle in the 1-d well, we can simply tack on a time dependent function to the spatial (time independent) solution. The temporal (time) function still depends on the energy levels.

For the 1-d well, the solutions are:

$$\psi(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$

We can tack on the temporal function: $e^{\frac{-iE_n t}{\hbar}}$

But the full solution to $\Psi(x,t)$ is a superposition of states.

So $$\Psi(x,t) = \sum_{n=0}^\infty a_n \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)e^{\frac{-iE_n t}{\hbar}}$$

And we can obtain the $a_n$ ("probability amplitudes") from the normalization condition:

$$\int_{-\infty}^\infty \mid \Psi(x,t) \mid^2 dx = 1 = \sum_{n=1}^\infty\mid a_n\mid^2$$

So in order to do this, we must be given initial conditions such as the behavior of $\Psi(x,t)$ at $t=0$. From the $\Psi(x,0)$ function we can find the $a_n$ and find how it evolves in time.

(2) Each eigenstate has quantized energy eigenvalues. The energy of the system is well defined but that does not mean the momentum is constant. Quantum mechanics is a probabilistic theory, and the momentum of a particle in a confined space is a probabilistic distribution, as is the position. You'll learn about Heisenberg's uncertainty principle, where the basic idea is that we cannot know anything about a particles momentum if it's position is completely known. In contrast, if the momentum is completely known, the position of the particle can be thought of as being spread out through all space; aka, we have infinite ignorance about the particle's position.

(3) Not entirely sure what is being asked here, but if you're asking if there is a way to determine an expectation value, then yes.

The 'distribution of the observable' can be evaluated as the expectation value of an operator, such as $x, \hat{p}, \hat{\mathcal{H}}$

Then the expectation value of $x$ will be:

$$\langle x \rangle(t) = \int_{-\infty}^\infty x~\Psi^*(x,t)\Psi(x,t)$$

This tells us the expected position (probabilistic) as a function of time, which is part of a distribution - all contained within the wave function. Essentially, the expectation value follows this probabilistic distribution.

All information about a particle is contained in its wave function $\Psi(x,t)$, but the only measurements of an operator which we can actually observe is one of its eigenvalues.


Edit: (1) There are some instances when the coefficients are all zero aside from a set number of states, such as $$\Psi(x,t) = \sum_{n=1}^2 \frac{1}{\sqrt{L}}\sin\left(\frac{n\pi x}{L}\right)e^{\frac{-iE_n t}{\hbar}}$$ If $\Psi(x,0)$ was completely unknown, we cannot determine the time evolution of the wave function without first measuring a state. The time dependent Schrödinger equation is an initial value partial differential requiring us to specify a state, usually $t = 0$. But perhaps the time independent equations could be of use. It would be easy to assume the form of the wave function as a free particle: $\psi(x) = Ae^{ikx}$. Otherwise, the form of the wave function is dependent on the potential containing it (inf. sq. well, harmonic, finite well, etc).

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  • $\begingroup$ You don't get the $a_n$'s from normalization. You get them from integrating $\Psi(x,0)$ against your basis functions (in this case, the energy eigenfunctions). As far as (3): the expectation value is not the same as the "distribution of the observable". To get that distribution, you have to expand the wave function in the eigenbasis of the corresponding observable. For instance $\Psi(x)$ is already expanded in the position basis, and so $|\Psi(x)|^2$ is the distribution for $x$. Your $a_n$'s compose the distribution for $\hat{H}$. And so on. $\endgroup$ – march Oct 25 '16 at 23:27
  • $\begingroup$ Sorry yes I was a bit vague here. But the $a_n$ should arise from the normalization condition for the 1-d infinite well. According to the orthonormality relation, $$\int_{0}^L \psi_{n}^*(x)\psi_n(x) dx = \delta_{n,m}$$ where the kronecker delta function is 0 for n not equal to m, and 1 for n = m. For the 1d infinite well, this leads to $$\int_{-\infty}^\infty \mid \Psi(x,t) \mid^2 dx = \sum_{n=1}^\infty \mid a_n \mid^2$$, where $$\sum_{n=1}^\infty \mid a_n \mid^2 = 1$$ $\endgroup$ – bleuofblue Oct 26 '16 at 0:59
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I am trying to answer point 3. of your question. The answers to the other points have been correctly given by bleuofblue. If you have a wave function $\psi$ and an observable $a$ with its corresponding operator $A$ which has a complete set of normalized orthogonal eigenfunctions $\psi_n$ with corresponding discrete eigenvalues $a_n$, then you also get the probability distribution of possible measurement values of the observable a. The probability distribution for the measurement values $a_n$ is given by $$P_n(a_n)=|\langle\psi|\psi_n\rangle|^2$$ The expectation value is given by $\langle a\rangle=\langle\psi|A\psi\rangle$ and the standard deviation of the distribution is given by $\sqrt{\langle\psi|A^2 \psi\rangle-\langle \psi|A \psi\rangle^2}$ Similar formulae exist for continuous spectra of eigenvalues.

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  1. Fermi Dirac statistics bridges the gap between thermodynamics and quantum mechanics and can be used along with the schrodinger equation to determine the probabilities of the various bound energy states, which in turn can be used to determine the coefficients of the general wavefunction. The extra measurement necessary for this determination, as in the classical case, is temperature. This kind of an experiment shows up regarding electron occupation states in a nanoscale semiconductor, which can be approximated by the particle-in-a-box scenario under certain conditions.

Links that I found helpful in understanding the physics (ymmv):

$\bullet$ Don't electrons have discrete energy levels in semiconductors?

$\bullet$ http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-701-introduction-to-nanoelectronics-spring-2010/readings/MIT6_701S10_part1.pdf

$\bullet$ http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-701-introduction-to-nanoelectronics-spring-2010/readings/MIT6_701S10_part2.pdf

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protected by Qmechanic Oct 26 '16 at 8:42

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