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(Kinetic Energy) Operator as a Matrix

by ylem
Tags: energy, kinetic, matrix, operator
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ylem
#1
Oct24-07, 11:44 AM
P: 32
Hey! I am completely confused...

In quantum mechanics at University we are learning about operators. I don't understand what you actually have to do to write an operator as a matrix...

In my notes, I'm looking at the kinetic energy operator in the Hamiltonian...

I can see for something basic - like an operator that makes things the number 3 when you have an orthonormal basis set that the matrix is simply

3 0 0
0 3 0
0 0 3

But I don't know how to adapt this to something more complex...

Thanks for your help! I hope I posted this in the right place :-s

Sam x
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CompuChip
#2
Oct24-07, 01:22 PM
Sci Advisor
HW Helper
P: 4,300
Sometimes, you can write an operator as a matrix. Note that, just like in ordinary linear algebra, the entries of the matrix depend on the basis you are in. You shouldn't forget, a state is a vector in a (Hilbert) space, but the vector looks different in different bases you choose (in analogy with |R^3: the vector (1, 1, 0) in Cartesian coordinates and (1, pi/2, 0) (I think) in spherical coordinates are the same vector, though the components look very differently).

So for example, if I write the state of a spin system in the basis {spin up, spin down} along the z-axis, I can write the matrix of the spin z-component as
[tex]\hat S_z = \begin{pmatrix}\hbar/2 & 0 \\ 0 & -\hbar/2 \end{pmatrix}.[/tex]
But when I take another basis (for example, the spin along the x axis) the matrix of the same operator will look different.

Also, some operators are infinite. For example, the position operator in one dimension has infinitely many eigenvalues (namely, one for each position x). So if we apply the [itex]\hat x[/itex] operator to a state [itex]\Psi[/itex], it will give us the position. Now we can again choose a basis for [itex]\Psi[/itex], consisting of all positions x on the real line. The "components" of [itex]\Psi[/itex] are usually written [itex]\langle x | \Psi \rangle[/itex] or [itex]\Psi(x)[/itex] (it is a vector, but it's components are not labelled by some i = 1, 2, ..., n; but by a continuous index x. Though we're looking at it as a vector, such a thing is more commonly called a function). We can write the operator as a(n infinite) "matrix", namely the diagonal matrix with all the (uncountably many!) eigenvalues x on the diagonal. But if we choose another basis, this "matrix" will look quite different. For example, if we choose as basis the momentum p, the x-operator will look like [itex]i\hbar \frac{\partial}{\partial p}[/itex] in that basis and the matrix will no longer be diagonal (in fact, it is not possible to choose any basis such that the x and p operator matrices are diagonal at the same time, which have severe consequences with physical relevance).

So I'm really tired, but I hope I didn't write too much nonsense here
BerryBoy
#3
Oct24-07, 04:28 PM
P: 183
OK, so let me first just state that the energy eigenvalues for a particle in an infinite potential well with width [itex]2a[/itex] is as follows:

[tex]E_n = \frac{\hbar^2 \pi^2 n^2}{8ma^2}[/tex]

And we know that the associated eigenwavefunctions for the system are:

[tex]\psi_n(x) = \frac{1}{\sqrt{a}} \sin (\frac{n \pi x}{2a})[/tex]

Now if we have a state that contains different amounts of these wavefunctions we can write the state as a coloumn vector of eigenwavefunctions:

[tex]\Psi_A(x) = \left(\begin{array}{cc}1\\0\\-1\end{array}\right)[/tex]
[tex]\Psi_A(x) = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{a}} \sin (\frac{\pi x}{2a}) -\frac{1}{\sqrt{a}} \sin (\frac{3 \pi x}{2a})\right)[/tex]

So now recall the definition of an operator:
[tex]\hat{O}\psi = \lambda \psi[/tex]

So if I use the Hamiltonian operator, I'd expect it to pull out the energy eigenvalues for the states present. So I can write this as an arbitrary dimension matrix:

[tex]\frac{\hbar^2 \pi^2}{8ma^2} \left(\begin{array}{cccc}1&0&0&0\\0&4&0&0\\0&0&9&0\\0&0&0&n^2\end{array }\right)[/tex]

So that we can now write something that is easy to work with:

[tex]\hat{H}\left(\begin{array}{c}\psi_1\\\psi_2\\\psi_3\\\psi_n\end{array}\ right) = \frac{\hbar^2 \pi^2}{8ma^2} \left(\begin{array}{cccc}1&0&0&0\\0&4&0&0\\0&0&9&0\\0&0&0&n^2\end{array }\right)\left(\begin{array}{c}\psi_1\\\psi_2\\\psi_3\\\psi_n\end{array} \right)[/tex]

Does that help?? I'm not sure which part of the writing the operator as a matrix you didn't understand so I was very vague on details (just gave an example as an addition to CompuChips post, but worked on the other side of the equation). I think that writing matrices in calculations just simplifies equations that are in many orthogonal states. Phew, LATEX is fun :S

CompuChip
#4
Oct26-07, 12:29 PM
Sci Advisor
HW Helper
P: 4,300
(Kinetic Energy) Operator as a Matrix

Quote Quote by BerryBoy View Post
Phew, LATEX is fun :S
It is. You can even just do something like
[tex] \begin{pmatrix}
a & b & c \\
d & e & f \\
\vdots & \ddots & \vdots \\
x & \cdots & z
\end{pmatrix}[/tex]
(note that there is no array, no column specification, and no manual bracing)
Oh and \vdots and \ddots will make your matrices even more beautiful

On topic, I think you are right the TS should try to ask a more specific question now for more details.
ylem
#5
Oct29-07, 05:00 AM
P: 32
Thanks so much for your help! I think I understand it a bit more now! :-) x


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