I'm trying to get to grips with the Schrödinger equation by looking at a free particle. I'm certain at some point I massively misunderstood something.

According to a textbook and a lecture the free particle moving in positive x direction can be described by

$$ \Psi(x,t) = A e^{i(kx - \omega t)} = \psi(x) \cdot e^{-i\omega t} $$

Classically, I would expect the particle to move along its path with some constant velocity $v = (x-x_0)/t$, so I would like to determine the probability $P(x,\Delta x,t)$ to find the particle between $x$ and $\Delta x$ at time $t$ in order to compare it with the classical location $x(t) = x_0 + v\cdot t$.

Since I'm looking for a location, I have to use the (trivial) location operator $\hat{r} = r$ and I get:

$$ P(x,\Delta x,t) = \int_x^{x+\Delta x}\Psi(x,t)^* \hat{\Psi(x,t)} dx\\ = \int_x^{x+\Delta x}\Psi(x,t)^* \Psi(x,t) dx\\ = A^2\int_x^{x+\Delta x} {e^{i(kx - \omega t)}}^* {e^{i(kx - \omega t)}} dx \\ = A^2\int_x^{x+\Delta x} dx = A^2 \Delta x $$

Which doesn't make sense to me at all, since it doesn't depend on either $x$ or $t$. The particle is not at all locations along the x-axis with the same probability at all times; I would rather expect it to move along the axis with the velocity $v$ but regardless which constants I shove into $A$ by using normalisation constraints, $P$ will never depend on $t$. But according to my understanding it should.

Obviously, my understanding is wrong and/or I made some mistakes in my calculation. Where am I going wrong?

  • $\begingroup$ This doesn't work for the free particle, since the wave function isn't square integrable: $\int 1 dx = \infty.$ But do repeat this exercise for a particle in a box, for example! $\endgroup$ – Vibert May 1 '13 at 9:00
  • $\begingroup$ @Vibert: If I trap the particle, this merely changes the normalisation and gives me a finite value for A. But the core derivation doesn't seem to change. $\endgroup$ – bitmask May 1 '13 at 9:05
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    $\begingroup$ By the way, you aren't inserting the position operator but the operator that is a "box function" which is 1 on the interval $[x,x+ dx]$ and 0 otherwise. The expectation value of the position is not defined: $\langle x \rangle = \int x \psi^* \psi = \int x = \infty.$ $\endgroup$ – Vibert May 1 '13 at 12:45

When you solve the Schrodinger equation for a free particle you get a family of solutions of the form $\Psi(x,t) = A e^{i(kx - \omega t)}$ and all superpositions of these functions. So just solving the Schrodinger equation doesn't give you a solution for a specific particle. For that you need to specify the initial conditions.

If you take the solution to be $\Psi(x,t) = A e^{i(kx - \omega t)}$ then you are (without realising it) specifying the initial condition to be a completely delocalised particle i.e. one for which we have precise knowledge of the momentum but no knowledge of the position. That's why when you attempt to calculate the position you get a silly answer.

If you specify the initial conditions as $\Psi(x, 0)$ then you have effectively created a wavepacket describing your particle, so it does have a finite uncertainty in position, and of course now a finite uncertainty in momentum. You can now calculate the expectation value of position as a function of time.

Your $\Psi(x, 0)$ will probably be expressed as a linear superposition of the plane wave solutions. To calculate the superposition just Fourier transform your $\Psi(x, 0)$.

Response to comment:

In your comment you ask:

I would have to plug in some constraints first and get a value for A or how would I go about doing that?

but it isn't just a matter of choosing the value of $A$ in $A e^{i(kx - \omega t)}$ because this doesn't describe a localised particle whatever value you choose for $A$.

Suppose at time $t = 0$ we know the position of the particle precisely, $x = 0$. This means the initial wavefunction is a delta function:

$$ \Psi(x, 0) = \delta(x) e^{-i\omega t} $$

i.e. $\Psi(x, 0)$ is zero everywhere except at $x = 0$. The position of this particle is obviously $x = 0$.

The trouble is that it isn't obvious how this wavefunction evolves in time. We know how the plane waves evolve in time, so we can easily calculate the time evolution if we could express the $\delta(x)$ function as a sum of plane waves:

$$ \delta(x) = \sum\limits_i A_i e^{i(kx - \omega t)} $$

The problem is working out how to do this sum, i.e. what are the values of the coefficients $a_i$ and how many terms we need in the sum. We can work this out by Fourier transforming our $\delta$ function, because this is exactly what a Fourier transform does. It expresses any function as an integral of plane waves. The Wikipedia article I've linked goes into more detail on this. The answer is that:

$$ \delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i(kx - \omega t)} dk $$

In fact choosing a $\delta$ function as the initial conditions isn't helpful because if we have an exact position we have infinite uncertainty in momentum, and if the momentum is infinitely uncertain we can't calculate the future position. If you're trying to describe a real system you would choose something like a Gaussian:

$$ \Psi(x, 0) = k \space e^{-(x^2/\Delta x^2)} $$

This describes a particle with the expectation value of $x = 0$ and the uncertainty in $x = \Delta x$. You can now Fourier transform your $\Psi(x, 0)$ to express it as an integral of plane waves, and you can now calculate the expectation value of $x$ as a function of time.

  • $\begingroup$ The first half of this answer actually makes sense to me but the second I don't quite get (and the downvote confuses me). Anyhow, to get a non-silly answer, I would have to plug in some constraints first and get a value for A or how would I go about doing that? $\endgroup$ – bitmask May 1 '13 at 9:44
  • $\begingroup$ The answer is correct, I dont get the downvote. In the last statement -- it will be a linear superposition of free waves, just to make it more clear. $\endgroup$ – Peter Kravchuk May 1 '13 at 10:02
  • $\begingroup$ @bitmask: the downvote confuses me too :-) I've edited my answer to respond to your comment. I hope this makes it clearer and not even more confusing :-) $\endgroup$ – John Rennie May 1 '13 at 10:09
  • $\begingroup$ It seems to me that you have explained as well as humanly possible and the issue is now a bit clearer to me, thanks. But it now kills the parts that I thought I had already understood ... back to the books, then :) $\endgroup$ – bitmask May 1 '13 at 10:25
  • $\begingroup$ @bitmask: take heart, this is a very common mistake for beginners to QM. Have a look at farside.ph.utexas.edu/teaching/qmech/lectures/node25.html for more info on the maths involved and demonstrations.wolfram.com/EvolutionOfAGaussianWavePacket for an animation. $\endgroup$ – John Rennie May 1 '13 at 10:35

I want to elaborate on John Rennie's answer. The Schrodinger equation for a free particle is ($\hbar=1$): $$ i\frac{\partial}{\partial t}\psi=-\frac{1}{2m}\frac{\partial^2}{\partial x^2}\psi. $$ It is a first-order differential equation in variable $t$. To solve it, you should specify initial data, say, $\psi(t=0)$. At this point, you should be aware that the uncertainity principle works, so you cant say that you have a particle travelling form $x_0$ with velocity $v$. If you choose a plane wave as your initial data, you are saying that the momentum(velocity) is exact, and therefore the uncertainty in $x$ is infinite -- the particle can be at any point. Such states do not occur in real experiments. However, you can choose some reasonable initial data like: $$ \psi(x,0)=A\exp(-x^2/4\Delta x^2+ipx). $$ If you take the $|\psi|^2$, you will see that it is a particle near $x=0$ (Gaussian distribution with dispersion $\Delta x$). If you go to momentum representation, you will see that momentum distribution is also Gaussian around $p$ with dispersion about $1/\Delta x$ (in accordiance with the uncertainty principle): $$ \psi(k,0)=A\sqrt{4\pi\Delta x^2}\exp(-(k-p)^2 \Delta x^2) $$

The question is, how to solve this equation with this initial data? The easiest way is to make a Fourier transform and write the initial data as the inverse Fourier transform: $$ \psi(x,0)=\int \frac{dk}{2\pi} \psi(k,0)\exp(ikx) $$ You see that it is a superposition of plane waves, so, as far as the equation is linear, you can just solve it for plane waves. The solution the reads as: $$ \psi(x,t)=\int \frac{dk}{2\pi} \psi(k,0)\exp(ikx-i\omega(k)t),\\ \omega(k)=k^2/2m $$ You can take this integral because it is Gaussian and obtain: $$ \psi(x,t)=A\sqrt{\frac{2\Delta x^2}{2\Delta x^2+it/m}}\exp\left\{-\frac{(x-pt/m)^2}{4\Delta x^2+2it/m}+ipx-itp^2/2m\right\}. $$ You can see that your Gasussian distribution for $x$ is travelling with velocity $p/m$, just as expected. You can also note that the uncertainity in $x$ is growing.

  • $\begingroup$ Thanks! I had to read this a couple of times but I think I got the gist of it. It seems I failed horribly when choosing a simple example to understand the wave function :) $\endgroup$ – bitmask May 1 '13 at 11:27
  • $\begingroup$ But how is it related to the question about P(x,\delta x,t)? $\endgroup$ – freude May 1 '13 at 12:17
  • $\begingroup$ @freude $P(x,\Delta x,t)=\int_x^{x+\Delta x} |\psi(x,t)|^2 dx\simeq|\psi(x,t)|^2\Delta x$. Gaussian wavepacket is an example of an experimentally realizable wavefunction. $\endgroup$ – Peter Kravchuk May 1 '13 at 14:21
  • $\begingroup$ What does this probability mean? $\endgroup$ – freude May 1 '13 at 14:23
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    $\begingroup$ @freude. Sorry, I have to go. We are talking about basic principles of quantum mechanics. You can read about interpretation of the wavefunction in any QM textbook. If you think that you already understand this and I am wrong, then I am not likely to convince you right now. Thanks for the discussion. $\endgroup$ – Peter Kravchuk May 1 '13 at 14:35

The problems you have been encountered are related to that fact that you try to calculate probability of some unphysical situation. Quantum mechanics can give you probability of an outcome from some experiment. This wave functions does not contain any information (restrictions) concerning the way how you are going to measure it and what you are going to measure (which operator coordinate or momentum). In other words, you compute a probability of nothing. For example, to measure the particle's momentum, people can use double slit experiment as a measurement setup. If you want to measure a space coordinate, you have to localize particle first using screen with hole or something like this. For the interpretation of such experiments, the problem "particle in box" can be useful.

  • $\begingroup$ Well, the example is discussed in a (very good) textbook and a lecture (they just don't go into detail regarding the particle's position). So it must have some meaning. A free particle with a constant velocity doesn't strike me as an "unphysical" situation. I admit that I might have done some errors in my calculation, but the premise should be valid. $\endgroup$ – bitmask May 1 '13 at 9:39
  • $\begingroup$ In quantum mechanic, "a constant velocity" is already an unphysical situation. Everything is fluctuating. $\endgroup$ – freude May 1 '13 at 10:06
  • $\begingroup$ You did everything right in calculations. So, if initial statements are correct, you should be satisfied by what you have got, i.e. square of the norm multiplied by a volume of space. Have you heard something about observables? $\endgroup$ – freude May 1 '13 at 10:08

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