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Why is it so important in Quantum Mechanics to have a diagonalizability in order to have your set of vectors to span a space. Can you perhaps give me a proof that diagonalizability implies spanning of your vector space? (Griffths failed to show this)

About the relationship between diagonalization and spanning: Griffiths says on page 452 in Appendix A just below equation A.80 that “…a matrix is diagonalizable if and only if its eigenvectors span the space.” This is a feature of linear algebra in general, not just as applied to quantum mechanics. I or someone else could probably dig up a formal proof if you want, but think of it this way: each vector in a set that can't be expressed as a linear combination of the other vectors in the set represents a dimension. Your set of vectors spans the space if and only if the set of vectors describes the same number of dimensions as the space. The vectors in a diagnalized matrix clearly can't be expressed as sums of the other elements, and the number of elements is equal to the number of dimensions in the space.

If an nxn matrix A has less than n distinct eigenvalues, then A may or may not be diagonalizable. Whether or not it is diagonalizable is a question of whether there exists an eigenbasis for A. To find this we have to find the eigenvalues of A and corresponding eigenspaces. A is diagonalizable iff the dimensions of the eigenspaces add up to n. If we then form an eigenbasis S by concatenating the bases of these eigenspaces, the matrix D=(S^-1)AS is diagonal, and its ith entry is the eigenvalue associated with the ith eigenbasis.

Also, This is for all you Math crazy folks again. When does a partial derivative turn into a ordinary derivative? Is there a rule out there when this happens. A link would be fine on this.

Thanks! -It was the dog who killed the cat…-Schrödinger's Dog

I'm not sure about the first question, but a partial derivative is a derivative of a multidimensional function with respect to a single variable. This means you examine the function as only one variable changes. The total derivative of a multivariable function means taking the derivative, but allowing the other variables to change also. If f is a function of x and y, then the total derivative with respect to x would be: <math>\Large \frac{d f}{d x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y} \frac{\partial y}{\partial x}</math> For a function which depends only on a single variable, only the first term in the above equation exists.

Yuichi - The above relation makes sense when f depends on both x and y, but some additional consideration introduces a relation between x and y, i.e. <math>y = y(x)</math>.

I thought it should be just <math>\Large \frac{d f}{d x}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}</math>. There is no more <math>\frac{\partial f}{\partial x}</math> on the right side of the equal sign.

Yuichi - This reduction is part of what Poit0009 is saying but in a special case where f does not have an explicit dependence on x. i.e. f only depends on y and y depends only on x.

Hmmm, ok, but I don't see how this works in equation 1.21. When the derivative was outside the integral, it was ordinary, and when it came inside the integral, it became a partial derivative. They are plenty of other examples where Griffiths does this.

-“Cats & Dogs”-Schrödinger's Dog

Sch's Dog, in 1.21, it's because the derivative passes into the integral. This is legit if you look at limit laws (recall that derivatives and definite integrals are really limits). On the left, you do the integral before differentiating. You would end up with the derivative of a function of t, since you integrated with respect to x. The x's are gone - evaluated. On the right hand side, though, the derivative is inside, and applies to a function of more than one variable, so it's a partial derivative.

I think you are just thinking about this too hard. If the function you are differentiating is a function of one variable, it's a total derivative. If it's a function of more than one variable, it's a partial derivative. Griffiths also explains this in the paragraph immediately after 1.21.

Since momentum is a complex operator, does that mean that we never talk about the momentum itself in QM. <math> {p }={-ih }\frac{\partial }{\partial x}</math>. Same question for total energy <math> {E }={ih }\frac{\partial }{\partial t}</math>. Due to uncertainty principle, expectation value is the one that we care about? I interpret it this way intuitively, but was not sure about it. Anyone can come up with a more rigorous explanation?

We can talk about momentum/energy, but since there is an uncertainty to it, we have to describe it through an expectation value. This expectation value, tells us where most of the action is happening, in a statistical sense. In the book they say that you shouldn't talk about the expectation value as a “average”, but an “average repeated over and over again, with identical ensembles of particles,” but I take it loosely to be the “average”, and think in a sense, that the expectation value tells us what the momentum/energy in a non-classical way.

This may not be a rigorous argument though…give me a couple days…have to work on MXP :P.

Just because the operator is complex does not mean that it is not useful to us. The wave function itself is, in general, a complex function, but it contains all of the information that we can extract from the state, so it must be useful! An operator acting on the wave function represents the measurement of that state, and in order for these measurements to make sense, we need to get real values, not complex ones. You are right, it would not make sense to have complex values for the momentum of a state. So, in order to get real values, we require that all observables (operators) in quantum mechanics be hermitian; they must have real eigenvalues. The Hamiltonian, something that we are familiar with, is an operator that yields energy eigenvalues, and these are always real because we require the Hamiltonian to be hermitian. Also, the i in the momentum operator is there specifically to make it hermitian, as when you try to use the operator <math>\frac{d}{dx}</math> by itself, it is not hermitian, so it cannot be a possible observable.

In discussion, we looked at a particular wave function, and after normalization, obtained <math>A=\sqrt{\frac{3}{b}}</math>, if I recall. Our TA pointed out that it is actually more correctly <math>A=\sqrt{\frac{3}{b}}e^{i\theta}</math> (right?).

My question is this: how can we come to this result without knowing it ahead of time? How can this come out of the math?

I dont think he said <math>A=\sqrt{\frac{3}{b}}e^{i\theta}</math> is more correct…I think he was explaining that if you had a wave function with that normalization factor and you would square it the exponential part would cancel…

What I understood was that we make an assumption when we write <math>A=\sqrt{\frac{3}{b}}</math> that our function wasn't complex. It isn't necessarily “more” correct – because there is no reason to assume the solution should be complex as the part of the equation we were given didn't have any complex numbers. I think he was actually just showing that we should be aware of the assumption we made - and know that if we weren't wanting to make that assumption we should add it.

when <math> |z^2| = 1 </math>, the answer, if z is real, is <math> z = \pm 1</math> as you may be familiar with. If z is complex, the answer is <math>z={\rm e}^{i\theta}</math> where <math>\theta</math> is any real number. But if you don't know (remember) this answer already, you can think in the following way.

If you express z by <math>z=x+iy</math>, x and y are real, <math> |z^2| = 1 </math> will lead to <math> x^2 + y^2 = 1 </math>. This represents a unit circle centered at the origin in the complex plane. So if you express z in the polar (is this the right terminology?) coordinate (<math>z=r{\rm e}^{i\theta}</math>), the radius, r is 1, while there is no constraint on the polar angle and therefore, <math>z={\rm e}^{i\theta}</math>.

my question from chapter 1 is more about the definitions on page 4; it says “observations not only disturb what is to be measured, they PRODUCE it…” and that “the particle wasnt really anywhere” photons from the measurement device could disturb the particle we are measuring but do they really produce them in some cases also?

With regard to Andromeda's question about chapter 1, according to the Copenhagen interpretation of quantum mechanics, measurements do produce the states that you observe. Before the measurement, the state is in a superposition of all possible eigenstates that could be yielded from a measurement. After measurement takes place, the state collapses into one eigenstate. There is no way to tell ahead of time with certainty which eigenstate will be observed, unless the state was cooked up ahead of time, in which case a measurement of the same observable will, with certainty, yield the eigenstate that you prepared. Also, after making a measurement and collapsing the state, if you try to measure the state with an incompatible observable with respect to what you just measured, you will destroy the information of the state.

I have never understood why the probability density function is defined the way it is. Is it just an arbitrary choice to have the integral of the squared wave function be the probability density function or is there deeper reasoning behind it?

This may help: http://en.wikipedia.org/wiki/Probability_amplitude

Yuichi - 22:00 9/9/09

I think chavez is asking why it was sensible to speculate that <math>|\psi^2|</math> represents the probability density, and what theories and/or experiments support this speculation.

It would make sense that some sort of squaring would be necessary. <math>\psi</math> is a complex function, so some sort of operation should be required to find a physical probability.

I was questioning the same thing and went back to our freshman quantum book…section 5-3 of that book (Born's interpretation of the wave function) says: I quote : “Since the measurable quantity probability density is real and non negative, whereas the wave function is complex, it is obviously not possible to equate probability density to wave function. However, since wave function squared is always real and non negative, Born was not inconsistent in equating it to probability density.”

I was reading the same part of that text, Andromeda. Another part I found interesting here was: “Since the motion of a particle is connected with the propagation of an associated wave function, these two entities must be associated in space. That is, the particle must be at some location where the waves have an appreciable amplitude.” It goes on to say that <math>P(x,t)</math> must have an appreciable value where <math>\Psi(x,t)</math> has an appreciable value.

For probability density functions in general, there are two main requirements:

1. A probability density function <math>P(x)\geq 0</math> for all x.

2. <math>\int_{-\infty}^{\infty}P(x) = 1</math>. This means that p(x) has to go to 0 as x goes to infinity.

Since <math>|\Psi|^2</math> fulfills both of these (after normalization), it can be thought of as a probability density function. It doesn't appear that there is any direct evidence to say that it is the p.d.f. of the wavefunction, but there seems to be a lot of indirect evidence that pushes us towards that conclusion.

This isn't exactly what one might say was right onTOP of what chapter 1 discusses – but it's something I was thinking about. We mentioned the exponents with complex operator's that may and may not be discussed in this class as they are complex and don't always have hold in physical interpretations of energy and momentum etc. – but it seems like there are so many times we ignore the negative solutions or the complex solutions because they don't have any 'physical interpretation' in the real world. Then someone like Feynman or Durac comes along and applies the negative solutions to something else and voila. What mathematicians and physicists had ignored for so long was sitting right under their noses waiting for someone to find the positron or opposite time directions etc. What are we sitting on when we ignore the complex solutions to the wave equation? different types of space or time? — Physicists : “I found out why your chicken is sick – but it requires a spherical chicken in a vacuum”

I don't have enough command in quantum mechanics to say anything beyond the following: unless you can come up with an interpretation for complex eigenvalues, you are likely out of luck. I think it would be worth discussing in class though, as I would be interested in what a senior physicist, or somebody who just knows more than I do, would say.

A question regarding the uncertainty principle: on page 19, Griffiths writes that “a spread in wavelength corresponds to a spread in momentum,” since <math>p=\frac{h}{\lambda}</math>. Based on that equation, it seems to me that a larger spread in wavelength would correspond to a smaller momentum spread. So, as seen in Figure 1.8, a wavelength with a well defined position has a large uncertainty in <math>\lambda</math>, and it would seem, a small uncertainty in momentum. This clearly contradicts the uncertainty principle, so what am I overlooking?

As far as my understanding of the uncertainty principle goes, “The more precise a wave's position is, the less precise is its wavelength, and vice versa.” So in Figure 1.8 there is a well-defined position, and consequently a poorly-defined wavelength (since it isn't periodic). So I think it has a small uncertainty in position and a large uncertainty in momentum, which agrees with the conclusions the uncertainty principle makes.

Recall that the difference between a position distribution and momentum distribution is just a Fourier transform. Thus, you can think of a wide distribution in x-space corresponding to a thin distribution in p-space, and vice versa. You can think of it in this way or what the uncertainty principle dictates. Another situation to consider is the following: “Which state has the least uncertainty, i.e. least spread in position and/or momentum?” The answer is a Gaussian distribution, which makes sense when you try to apply the Fourier transform to it. I hope this answers your question.

There is an important distinction between the variables in the DeBroglie hypothesis to which you first refer, <math>p=\frac{h}{\lambda}</math> , and the Uncertainty Principle. The Uncertainty Principle states that you may not know two independent properties of a particle that fully define the particle's position and speed, and furthermore that you can only narrow the true values down to a certain minimum. The Uncertainty Principle relates the spread, or standard deviation, of variables whereas DeBroglie's formula relates the true values of the variables. The DeBroglie hypothesis represents momentum as a function of wavelength, <math>p(\lambda)</math>; any time there is uncertainty in wavelength, the error “propagates” through to momentum because the value and spread of momentum depends on the value and spread of the wavelength. While the nominal value of momentum varies inversely with wavelength, the “spread” of uncertainty varies directly. If you plug in simple numbers you can check this: if you know the wavelength is exactly–say–3, then p = h/3; however, if you can't narrow the spread of the wavelength to more than a value of “<math>\lambda</math> is somewhere between 2 and 4,” then your uncertainty in p is directly increased, and you can't know p more precisely than some value between h/4 and h/2. Therefore increasing uncertainty in <math>\lambda</math> increases uncertainty in momentum, p. (I hope that explanation made sense)

To go along with what Zeno was saying, I went ahead and used the error propagation formula:

<math>\sigma_{f(x, y)}^2=\sigma_{x}^2 \(\frac{\partial f(x, y)}{\partial x}\)^2 + \sigma_{y}^2 \(\frac{\partial f(x, y)}{\partial y}\)^2

p=\frac h\lambda

\sigma_{p}^2=\sigma_{\lambda}^2\(\frac{-h}{\lambda^2}\)^2</math>

As you can see, the uncertainty in p is proportional to the uncertainty in λ.