Quantum Mechanics

Inner or scaler products I think I understand but outer products are causing me difficulty, eg. |x><y|
where
|x> = a1|i> + a2|j> + a3|k>
<y| = b1<i| + b2<j| + b3<k|

where i,j,k are an orthonormal basis, I get that it's like a column vector by a row vector and forms a matrix but exactly how to multiply them I don't know.
if you're given the component form is there an easier way to evaluate it?

thanks! that's what I thought but wasn't sure if perhaps the non-diagonal entries were zero as a result of the orthogonal basis vectors. Is there an alternative way of expressing the outer product in terms of the components?

There's a question on a previous exam paper I'm looking at that asks to show that the eigenvalues of a unitary matrix are of the form exp(ia) where a is just a real number of some sort and I can figure out how to do it?

Cool, thanks, I read that proof in the book and understood it. It's the leap from knowing it's a complex number of modulus one to proving it could be written as exp(ia). Probably maths I should have known. Actually when I think about it it's really just Euler's formula. Thanks.

So I'm just wondering. Does quantum mechanics prove that matter is the creation of the mind and everything is just space of potential energy? By prove i mean that it can't explain the processes through experiment anymore thus a dead end... (sorry if i am using the word space incorrectly).

Iv'e been reading up on quantum theory and some of the findings are amazing. Though quantum mechanics is being analysed on a completely experimental/physical basis it seems that physical experiment has come to a dead end and something else has to take over to explain certain phenomena such as that double slit experiment and the idea that nothing is really there until we observe.

This is by far the most interesting thing i have ever come across and i feel it will finally tie the soul with the mind and thus discover the way to true happiness for everyone. (sorry if i sound too hippy-ish there but its true, we live to be happy)

I must admit i am quite impressionable and have probably been reading some crack-pot books=) I understand that i would need to learn a lot more about quantum mechanics before i can make any real deductions. I liked your example with the colours, made me think of superposition in a different way. But i still am curious about one or two things;

The particles that get measured in quantum mechanics are what exactly? Protons and neutrons or something smaller?

And the fact that these particles don't have any measurable position how does this effect the elements or matter it will form? "Is a gold bar not a gold bar when its smallest bits of matter are in a state of superposition"? And how come the observation by "us" concludes the position it will collapse into?

I know i'm out of my league here, i'll definitely read that book you mentioned and maybe get back to you in a few years;)

Thanks=)

Dooom wrote: »

If someone observes a particle, that person bumps the particle and disturbs it, or forces it to take just one of it's states, all from just observing it? And going on from there, how and why is it disturbed or forced to take a state just from looking at the thing?

To understand this, you first need to be clear what a quantum superposition is. Any quantum state is described completely by it's state vector (or equivalently it's wave-function). By this, I mean that all of the dynamic properties (position, speed, momentum, energy, etc.) are encoded in this vector. The state vector is the weighted sum of a set of orthogonal 'basis' vectors, each of which is orthogonal (at right angles) to each other. The weight of each vector in the sum is called it's amplitude. Squaring the amplitude gives the probability of measuring the system to be in this state. It is impossible to deterministically distinguish to quantum states which are not orthogonal.

I haven't yet said what the basis vectors are, and that's because we are free to choose. They could be positions, in which case the state vector would be written as a weighted sum of positions (i.e. |S>= a |0cm>+b|1cm>+...). Notice that there is no momentum in the state vector. Equally we could choose momentum as our basis, in which case we would be writing the state vector as a sum of momenta, with no position terms.

The important thing to note here is that a particular value of momentum corresponds to a superposition of positions, and a position corresponds to a superposition of momenta. This is where the uncertainty principle comes from, as it is impossible to be in a state which is simultaneously in a particular position and has a fixed momentum.

When we measure a particle, the result will be one basis vector, rather than a superposition. Thus when we measure something, we lose all the terms in the original superposition which are not consistent with the measurement. This is how measurement affects the state. It's not immediately obvious that you've affected the state, since the measurement out come must have been part of the superposition anyway, but it has a big affect when you then try to measure another property of the superposition, since you have already collapsed it.

Hope this helps.

Son Goku wrote: »

It is as of yet unknown what causes a particle to select one state in particular or in other words to lose the rest of the superposition. There are many different opinions on this, but quite a lot of people think it is an effect called decoherence, but it is unknown if that can explain it completely as of this point in time.

I can't say I completely agree with this. When you measure the state of a particle, your measurement aparatus becomes entangled with the particle. Unless this process is completely reversed (which is virtually impossible for most devices) the elements of the original superposition can no longer interfere. This is why you get one state and loose the rest. This does decohere the particle since the phases become spread over more than one particle. Hence if you neglect the second system you are left with a mixed state rather than a pure state in a coherent superposition.

Decoherence is a rather general term which describes any process in which the coherence of a superposition is eroded. Decoherence is caused by an interaction with other quantum particles in the environment, be this wanted, as in the case of measurement, or unwanted as in the case of noise in quantum systems.

The real argument is over why we percieve one particular state as the outcome, and on whether the other components of the superposition really exist. Why aren't we aware of superpositions directly? Why is there a prefered basis? And so on.

Son Goku wrote: »

I'm not advancing a particular opinion, all I'm saying is that there doesn't seem to be a general consensus across physics, although there is a general consensus within given areas of physics.

Certainly the question of what constitutes a measurement and the ontological status of other parts of the superposition is something of a philosophical question, and one which is very prominant in philosophy of physics. But that's not really what I'm getting at here. We understand exactly why measurement prevents us from getting inteference with other elements of the superposition.

Certainly the very first atom in your TEM tip is quantum mechanical, or the photon you use to observe a particle, etc. Once this interacts with the particle, you produce an entangled state of the two particles. Tracing out the particle used in the measurement device yields a mixed state, and so there is loss of coherence. Thus, unless you can actually undo this interaction, you cannot get interference effects from the original particle.

This is true even in the Copenhagen interpretation, once we consider a real apparatus. Why we get one value rather than another is really the measurement problem, not why we get eigenstates of the measurement operator.

Son Goku wrote: »

Unless I'm wrong I don't think there is a definitive answer as to why in a energy measurement for example you get one particular value and why this value corresponds to classical quantities and you have no contact with the the rest of the superposition that appears in the formalism.

See above.

Son Goku wrote: »

For instance for a contrasting view of things in Algebraic QFT people take the view that you get a definitive eigenvalue because eigenvalues are properties of events and a measurement is an event. Bohr thought QM "knew" about classical mechanics in some vague sense. Doesn't make much sense to me, but it just shows the radically different opinions people have on this. Even people who think it is all one big unitary operation are divided into Many-Worlders and the other camp (can't think of their name).

Again, these are views largely on why we get certain results, and are somewhat linked to human perception, rather than why we don't get interference.

Son Goku wrote: »

For instance the "Heisenberg cut" criticism of this:
would be, how does the apparatus know which state to entangle with and why is it a classical state? Why doesn't it entangle with a broad selection of states?

I'm not sure how an open quantum systems approach has a heisenberg cut criticism, but I'll try to answer your questions. The apparatus becomes entangled with the quantum system you are measuring in a way specificied by the interaction. For an Ising interaction, this will be by Z basis states picking up different phases, where as for an exchange interaction it will be a little more complicated, perhaps implimenting a controlled-NOT type interaction.

The states are only classical in that they are eigenstates of the measurement operator. They can quite easily be superpositions of positions, or momenta, or spin or what ever other property you are interested in. For example measuring photon polarization at different angles is measuring some particular superposition of horizontally and vertically (or clockwise and anti-clockwise) polarizations.

Son Goku wrote: »

Maybe I'm being naive, I've just gotten so many conflicting ideas from people and books. Not only is there no agreement of the answer to the problem, there is no agreement on what is the problem. There is a lot of people who still think entanglement with the apparutus doesn't uniquely answer why the rest of the superposition is lost.

Well, i suppose it depends what you've been reading. There is certainly a mystique surrounding measurement in quantum mechanics, and many people who express an opinion are often uninformed about the details of open quantum systems, for example. Pop sci books are often particularly bad at this.

The fact is, that if we are allowed to consider even one particle in the measurement apparatus quantum mechanically, we can fully explain why you don't get interference. We can also mimic this using single particles as the measurement apparatus (and actually do for quantum error correction).

I'll try to come back to the other points later, as I'm tight on time, but let me just try to give a quick answer to this question:

Son Goku wrote: »

I understand this, but a priori, if you get what I mean why are the eigenvalues classical quantities? This might sound vague, so let me be more concrete. Bohr and more recently Haag have pointed out that we write superpositions in terms of superposed "classical" states. For instance take a state |P> = a|E1> + b|E2>, where |E1> and |E2> are two states of definite energy. However at the quantum level there is just |P>, which could be expanded in any way. How is it that an energy measuring device can find the E-basis?
Basically why is it that if you want to do QM, you need to "already" know CM in a a priori sense? For instance I can understand your argument only in so far as we know classical measurment apparatus respond to a basis appropriate to them.

You get E1 or E2 when you measure the energy of the system, because you use the Hamiltonian for that system as the measurement operator. The Hamiltonian has specific eigenvectors with corresponding eigenvalues. If the system has a degenerate energy manifold (for example as is the case with the ground state of a ferro-magnet), then you simply project onto that vectorspace, but don't collapse superpositions within it. So you get E1 or E2 because they are inherantly encoded in the Hamiltonian of the system. Choosing a perturbed Hamiltonian, the new eigenstates could be |E1>+|E2> and |E1> - |E2>.

Don't worry, I know these aren't really trivial questions. I think perhaps you are just looking at quantum mechanics from different perspective, so perhaps my responses aren't terribly enlightening.

I'm wondering if perhaps I'm not understanding what you mean by a classical state. I presume you mean an eigen state of an operator which corresponds to an observable for which there is a classical equivalent. So for position, a classical state might is one which is a delta f uction in space. Is this correct?

If so, I think perhaps the confusion is steming from the fact that you can make any quantum state into a 'classical state' by choosing an appropriate operator for measurement. There is a whole formalism for describing quantum systems which has grown out of this idea, known as the stabilizer formalism, which leads to some pretty cool results about what systems we can simulate efficiently on classical computers.

I will hold off on answering your question, until I know what you mean, as I don't want to add to the confusion.

great unwashed wrote: »

Can any of you give me the best description of charge you've ever heard without mentioning quarks?

Charge is not intrinsically related to quarks. Quarks just happen to have a charge. Some leptons also have a charge, for example the electron.

Charge is simply a way of quantifying how particles interact with photons, of quantifying the strength of the electric field surrounding a particle.

great unwashed wrote: »

Is there an explanation for the exclusion priniciple - why these numbers of electrons and not others?

Yes, and it's actually very simple (although somewhat mathematical).

We call particles which have an anti-symmetric wave function Fermions. Particles with symmetric wave functions are called Bosons. Bare in mind that both these particles are identicle.

So, for fermions, a two particle state is
1/sqrt(2) {|psi_1> |psi_2> - |psi_2> |psi_1>}
while for a boson it is
1/sqrt(2) {|psi_1> |psi_2> + |psi_2> |psi_1>}
where |psi_1> and |psi_2> are the two states, one of which is occupied by each particle, and the position indicates which particle is in that state.

So, if the two particles are in the same state, then |psi_1>=|psi_2> and so the Fermions are in the state 0. This is not a valid state, since that probability of measuring the particle to be in any state is 0 (probabilities always add up to 1), and so it is impossible for this to ever happen. Hence the Pauli exclusion principle.

great unwashed wrote: »

When did physicists start talking about particle spin and what use does it have?

Particle spin is just the quantum equivalent of angular momentum of the free particle. Each particle is like a spinning top, and the rate at which is turns is it's spin. This is what gives rise to the magnetic field around charged particles with non-zero spin.

It was first proposed in 1925, and so has been around almost as long as quantum mechanics.

bogwalrus wrote: »

When you say in Quantum Mechanics "the observer is the observed" what does this mean?

I'm afraid I don't know what Bohm was refering to. He has a rather unconventional approach to measurement in quantum mechanics, so I'm not sure if he was simply elluding to a specific interpretation, or was trying to capture something about quantum mechanics in general. It's a nice sound bite, but not very informative.

Kareir wrote: »

I've read various books, but i was wondering if someone could explain the basic points of Quantum Mechanics to me.

Not really, it is a rather broad subject. If you have specific questions I would be happy to answer, but I can't really see anyone teaching you how to do quantum mechanics in full via the forum.

waitinforatrain wrote: »

1. The "observer". From what I understand, the photon begins to behave like a particle when it is being measured. Therefore the "observer" changes the thing observed. But does this observer need to be a conscious entity? Does the photon counter count as an observer? Does the photon counter have a discrete state and value before I (the experimenter) observes it?

No, it's just an effect of open systems (systems which interact with a broader system). If a system is closed, the evolution of the wave function is unitary, but when it is allowed to interact with other systems it causes it to appear as if the wave function collapses.

So the observer is simply another system. It doesn't need to be an animal or person, simply a few electrons will do.

waitinforatrain wrote: »

2. How difficult would it be to replicate the double slit experiment for the average joe? This is something I've found surprisingly difficult to find out.

Well, you can do the Young's slit experiment fairly easily, but doing so with single photons is much harder.

waitinforatrain wrote: »

3. How can we be sure that only one photon is firing at a time?

Usually an attenuated laser is used, but single photons are detected. Another way to do it is to use a single photon source. There are plenty of candidates which work. As it happens, though, we can do far better than simply using photons. The double slit experiment has been done with neutrons, atoms and even molecules.

waitinforatrain wrote: »

Ah thanks, that really clarifies it for me.

A few more quick questions for you if you get the time: What do you mean by 'unitary' (I googled and all I got was "Having a magnitude of one" which doesn't help me to understand it). Are you just saying that the wave function behaves as a wave, but when another system depends on it, it converges to a discrete value.

Is this 'wave' function vibrating in physical space, or is it a representation of the probability of the particle being at a certain point when observed?

You say "causes it to appear", does it not actually collapse or is it just relative to the system that is observing/interacting with it? Could two different systems measure it as collapsing to a different point?

It's a little more tricky. Basically quantum systems in pure states are systems in a state such that we can write the state as a superposition of classical states. By a superposition I mean that each classical state psi_i is waited by some complex amplitude A_i, such that |A_i|^2 is the probability of measuring the system in that state, so the 'wave function' which is just a list of these amplitudes has nothing to do with physical vibrations in general. Unitary operations are operations which preserve this structure.

When a quantum system interacts with the environment it undergoes a non-unitary operation. The effect of this is to bring the system into a mixed state. A mixed state is simply a classical probability distribution of quantum states. When the interaction is strong enough, the system will loose the complex phase of A_i (called dephasing), essentially making the system classical.

waitinforatrain wrote: »

Do you have any book recommendations for this stuff (hopefully that avoids complex maths)?

The Feynman Lectures, perhaps.

Azelfafage wrote: »

Put an Irishman in a Box-room with a bottle of whiskey.

Is he drunk or sober before you open the door?

That shows you the sheer absurdity of Schrodinger's "experiment".

Look, Azelfafage, at this stage you're comments are degenerating into complete nonsense (and frankly offensive nonsense). If you want to go start denying the existence of quantum mechanics, fine, but if so you may want to stop using pretty much every modern piece of technology, including your computer.

Azelfafage wrote: »

Schrodinger was merely making an analogy about Quantum Physics.

He was astounded that anybody believed about the cat.

A cat has too many atoms to be relevent to Quantum Physics.

That's not correct. The number of atoms has nothing to do with whether quantum effects will be visible or not. Decohrence (and hence the transition to classical physics) is entirely determined by interaction with an environment.

onedoubleo wrote: »

So I have two questions, is there anywhere else in nature we are seeing quantum effects like tunneling play a major role? And other than verification of theory is there any other reason to get large scale object to show wave phenomenon?

Q1) Photosynthesis, your sense of smell and bird navigation are all believed to exploit quantum effects.

Q2) To build a quantum computer.