Monday, November 15, 2004

Misconception of the Heisenberg Uncertainty Principle

One of the common misconception about the Heisenberg Uncertainty Principle (HUP) is that it is the fault of our measurement accuracy. A descripton that is often used is the fact that to observe the position of an electron, for example, one needs a probe, such as a photon, with very short wavelength to get any reliable accuracy. But a very short wavelength photon has a very high energy, and thus, the act of position measurement will simply destroy the accurate information of that electron's momentum. Hence, this is an example of the HUP.

While this is true, it isn't really a manifestation of the HUP. The HUP isn't about a single measurement and what can be obtained out of that single measurement. It is about how well we can predict subsequent measurements given the identical conditions. In classical mechanics, if you are given a set of identical conditions, the dynamics of a particle will be well defined. The more you know the initial position, the better you will be able to predict it's momentum, and vice versa.

The most direct way to illustrate this is using a single-slit measurement. Let's first consider the classical case so that we know what we normally expect to happen. I suggest you sketch the setup along (since I'm unable to show sketches on here, but will try to make an effort to clearly explain the orientation of things).

Let's say you have a source of classical particle that emits this particle one at a time on demand, and emits it with a constant velocity and kinetic energy. At some distance from this source is a single slit. For clarity sake let's say the slit is alligned along the x-direction, so that the width of the slit is along the y-direction. The orientation of the x and y coordinate axes is in such a way that (using the right-handed coordinate system) the z-axis is along the direction of propagation of the particles. So the direction of the z-axis is from the source to the slit, and beyond.

Beyond the slit is a screen of detectors (could be a photographic plate, a CCD, etc.) This detector records where the particle hits after it passes through the slit. Let's say this screen is a distance L after the slit.

Now, let's get some basics out of the way:

1.If a particle gets through the slit, then I can say that my knowledge of the position of the particle at the slit has an uncertainty equal to the width of the slit. Thus, the width of the slit Delta(y) is the uncertainty in the postion of the particle when it passes through the slit.

2. The y-component of the momentum of the particle can be found by looking at how far the particle drifts along the y-direction when it hits the screen. This makes the explicit assumption that no external forces acts on the particle at and after it passes through the slit, so that it's momentum remains constant from the slit to the screen (which is a reasonable assumption). Let's say the particle drifts from the center, straight-through line and hits the screen at a distance Y. If it takes the particle a time T to reach the screen (which we can assume to be a constant if screen distance from the slit is much larger than the width of the slit (i.e. L >> Delta(y)), then the y-component of the momentum is p_y prop. Y/T. Now, there is a measurement uncertainty here in determining where exactly the particle hits the detector. This measurement uncertainty depends on the resolution of the detector, how fine is the "mesh", etc. But this is NOT the "uncertainty" that is meant in the HUP. We haven't gotten to the uncertainty of the momentum YET. All we have is a measurement of the y-component of the momentum of the particle.

Now, let's do the experiment with the classical particles. You shoot the particles one at a time and record where it hits on the screen. Ideally, what you will end up on the screen is only one spot where the particles that pass through the slit hit. However, closer to reality is that you end up with a gaussian distribution at the slit, where the peak lies directly along the straight-through direction that has zero y-component of momentum. The uncertainty in the momentum then corresponds to the width of the gaussian distribution (full width at half maximum). Now THIS is Delta(p_x) as refered to in the HUP!

Let's make the width of the slit smaller. This means Delta(y) is smaller. You are now letting a smaller possible angle of incidence of the particle from the source to get through the slit. This means that there will be a smaller spread that is detected on the screen. The gaussian distribution will be thinner. So classically, what we expect is that as Delta(y) gets smaller, Delta(p_y) also correspondingly becomes smaller.

This is what we expect in classical mechanics. If all the initial conditions remans identical (I have the same source), then the more I know where the object is at any given instant, the more I can predict its subsequent properties. I can say what its y-momentum will be with increasing accuracy as I increase my certainty of its position by decreasing the width of the slit. I can easily predict where the next particle is going to hit since I will know what its momentum is going to be after it passes through a very narrow slit. My ability to predict such things increases with decreasing slit width.

Fine, but what happens with a quantum particle such as a photon, electron, neutron, etc.?

We need to consider two different cases. If the slit width is considerably larger than the deBroglie wavelength (or in the case of a photon, its wavelength) of the particle, then what you have is simply the image of the slit itself. The ideal situation would give you simply a "square" or gaussian distribution at the screen of the intensity of particles hitting the detector. This is no different than the classical case.

It gets interesting as you decrease the slit. By the time the width of the slit is comparable to the deBroglie wavelength, something strange happens. On the screen, the spread of the particles being detected start expanding! In fact, the smaller you make the slit width, the larger the range of values for Y that you detect. The "gaussian spread" now is becoming fatter and fatter. This is the single-slit diffraction pattern that everyone is familiar with.

Now THIS is the uncertainty principle at work. The slit width, and thus Delta(y) is getting smaller. This implies that Delta(p_y) is getting larger. Take note that the measurement uncertainty in a single is still the same as in the classical case. If I shoot the particle one at a time, I still see a distinct, accurate "dot" on the screen to tell me that this is where the particle hits the detector. However, unlike the classical case, my ability to predict where the NEXT one is going to hit becomes worse as I make the slit smaller. As the slit and Delta(y) becomes smaller and smaller, I know less and less where the particle is going to hit the screen. Thus, my knowledge of its y-component of the momentum correspondingly becomes more uncertain.

What I am trying to get across is that the HUP isn't about the knowledge of the conjugate observables of a single particle in a single measurement. I have shown that there's nothing to prevent anyone from knowing both the position and momentum of a particle in a single mesurement with arbitrary accuracy that is limited only by our technology. However, physics involves the ability to make a dynamical model that allows us to predict when and where things are going to occur in the future. While classical mechanics does not prohibit us from making as accurate of a prediction as we want, QM does! It is this predictive ability that is contained in the HUP. It is an intrinsic part of the QM formulation and not just simply a "measurement" uncertainty, as often misunderstood by many.


Tuesday, November 09, 2004

So You Want To Be A Physicist - Part 7

We are still stuck in the discussion of your fourth and final year of college. This time, I feel that a clearly explanation of the US graduate school system is warranted, especially to others from the rest of the world who intend to continue their graduate education in the US. This is because there is often a great deal of confusion, from the conversation that I've had, regarding what is required to apply for a Ph.D program in physics in the US.

The broad dichotomy of higher education in physics in US institutions can be lumped into (i) undergraduate education and (ii) graduate education. When you have completed your physics undergraduate education, you typically earn a degree of Bachelor of Science (B.Sc). (There are some schools that actually award a Bachelor of Arts in physics, but that's a different path that we won't discuss here.) This is what we refer to as your undergraduate degree. If you do decide to go on to graduate school, then the two different physics degrees available to you are the Masters of Science (M.Sc) and Doctor of Philosophy (Ph.D).

Now the next step is where US institutions differ from many educational system throughout the world. If you intend to pursue a Doctorate degree in physics, you do NOT need to first obtain a M.Sc. degree. Practically all of the universities in the US that I'm aware of require that you have an undergraduate degree to apply to a Ph.D program. Your undergraduate degree and transcripts of your undergraduate class grades are the ones being used to evaluate your candidacy. This is different from, let's say, the UK system, where you first pursue your M.Sc, and then after completing that, go on with your Ph.D. In US institutions, if you are pursuing your Ph.D, you can get your M.Sc "along the way", since at some point, you would have fulfill the requirements for a M.Sc degree. In fact, I know of a few people who didn't even bother declaring for their M.Sc degrees. So you will see some people with academic credentials as "B.Sc in physics from so-and-so; Ph.D in physics from so-and-so", with the M.Sc degree missing.

These differences have created a sometime confusing discussion from people intending to enroll in the graduate program in the US. The first confusion comes in when they check the average length of time to complete a physics Ph.D. Most are shock that the average length of time to complete a Ph.D in the US is 5 1/2 to 6 years. I was told that it takes an average of 3 years in the UK. However, if you consider what I have mentioned earlier, the length of time for a Ph.D is taken from the enrollment into the program by someone with a B.Sc degree, whereas the UK number is taken from the start of the program after someone has obtained a M.Sc. or equivalent. It takes an average of about 2 years to complete a physics M.Sc in the UK, I think. So now there is a explanation of the apparent discrepancy between the length of time. The total length of time to obtain a Ph.D after someone has a B.Sc degree is still roughly similar in both educational systems.

The second, of course, is the idea that one must have a M.Sc degree before applying for a Ph.D degree. Again, if one were to browse through the requirement for acceptance into a Ph.D program at a US institution, one will see that the major requirement is an undergraduate degree. I know of many international students who either (i) stayed in their home countries to get their M.Sc and then apply for a Ph.D program in the US, or (ii) apply explicitly for a M.Sc program in the US even though their goals are to obtain a Ph.D, because they assume that one must obtain a M.Sc first, before going on to a Ph.D program. This can actually create additional annoying problems, because one sometime has to REAPPLY for enrollment into the Ph.D program (this means you may have to pay again the application fee, fill in application forms, etc...) They also must apply for a change of status on their visas, because they are now pursuing a different degree.... In other words, these are all messes and annoyances that could have been avoided had one understands the graduate school system.

So remember: check the requirements for admission into a Ph.D program for a US institution. A B.Sc degree is required, not a M.Sc. So if you intend to pursue a Ph.D, apply directly for a Ph.D program, using your B.Sc. degree.