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Simple quantum problem - find eigenvalues, probabilities, expectation value?

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jeebs
#1
Sep14-10, 05:30 PM
P: 326
hi,
not strictly homework as my course doesn't get going again for a couple of weeks yet, but suppose I have a system with quantum number l=1 in the angular momentum state

[tex] u = \frac{1}{\sqrt{2}} \left(\begin{array}{cc}1\\1\\0\end{array}\right) [/tex]

and I measure Lz, the angular momentum component along the z-axis. The problem I am attempting is to find out the possible results and their probabilities, also the expectation value.

So, I got the operator matrix [tex] L_z = \hbar \left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&-1\end{array}\right) [/tex]

and after using that in the eigenvalue equation, [tex] L_zu = \lambda u [/tex] I multiplied the matrices together to find that [tex] L_zu = \frac{\hbar}{\sqrt{2}}\left(\begin{array}{cc}1\\0\\0\end{array}\right) = \frac{\lambda}{\sqrt{2}}\left(\begin{array}{cc}1\\1\\0\end{array}\right ) [/tex].

from this I see that the possible eigenvalues (ie. the possible results of measurement, right?) are [tex] \lambda = \hbar [/tex] and [tex] \lambda = 0 [/tex]. If I'm not mistaken, this is a sensible result given that the angular momentum quantum number l=1.

however, i'm not sure how to approach getting the probability of each outcome. I have this expression in my notes, [tex] P_a = |<a|u>|^2 =\left| \sum a_i^*u_i \right|^2 [/tex]. What I thought about this was maybe that the state |u> was supposed to be taken as a superposition of two basis states, one for each eigenvalue, 0 and [tex] \hbar [/tex]. So, if I named these basis states as a1 and a2, when I again used the eigenvalue equation I found:


[tex] L_za_1 = \hbar \left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&-1\end{array}\right) \left(\begin{array}{c}a11\\a12\\a13\end{array}\right) = \hbar\left(\begin{array}{c}a11\\a12\\a13\end{array}\right) [/tex] which leads to [tex] a_1 = \left(\begin{array}{c}1\\0\\0\end{array}\right) [/tex]

When I do the same thing again I find that [tex] a_2 = \left(\begin{array}{c}0\\0\\0\end{array}\right) [/tex].

This doesn't seem sensible to me, because how can that be combined with the eigenstate a1 in a summation that produces the state u? it cannot, as every element is zero and u has two non-zero elements, so I have given up at this point. I was looking at the coefficient ai in the probability expression above, but I wasn't sure how to actually work out what they were spposed to be.

Does anyone have any ideas what I am supposed to do here? Am I even remotely trying to do the right thing?
Thanks.
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diazona
#2
Sep14-10, 07:47 PM
HW Helper
P: 2,155
For one thing, your notation is quite inconsistent. You use u for both the given state and the eigenvectors, and then you use a for both the coefficients and the eigenvectors, so I'm not surprised you're getting confused on that basis alone.

Here's what I'd recommend: First fix up your notation. Let the state of the system be denoted by x, instead of u. So you have a state
[tex]x = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}[/tex]
Let the eigenvectors of Lz be denoted by um, where m ranges from -1 to +1, corresponding to the directional angular momentum quantum number. Note that if you wanted, you could make m run from 1 to 3 instead, or 4 to 6, or whatever - it's just that -1 to +1 is convenient because that way the number directly corresponds to the z-angular momentum of the eigenstate (in units of [itex]\hbar[/itex]).

Now that the notation is cleared up, you'll first need find the eigenvalues and eigenvectors of the operator Lz. This computation has nothing to do with any particular physical system, or with any particular physical state. In particular, you should not be plugging the given state x (which you had labeled u) into the eigenvalue/eigenvector equation. In fact, the given state is not even an eigenvector of the z-angular momentum operator, so the results you got out of that calculation are effectively nonsense. (By sheer luck, you happened to pick two of the correct eigenvalues)

To find the eigenvalues and eigenvectors, use the usual mathematical procedure: rearrange the equation to
[tex](L_z - \lambda_m) u_m = 0[/tex]
where both λm and um are unknowns, and find the determinant of the characteristic matrix [itex]L_z - \lambda_m[/itex]. (Or of course you could just plug the matrix into Mathematica or something) You will get a set of three eigenvalues and a set of three eigenvectors. Note that none of the eigenvectors will be totally zero - the zero vector is by definition excluded from being an eigenvector. When you try to find the eigenvector corresponding to eigenvalue 0, look for a solution in which not all the elements of the vector are zero.

At this point, you should decompose the given state x into a linear combination (superposition) of the eigenvectors.
[tex]x = a_{-1} u_{-1} + a_{0} u_{0} + a_{1} u_{+1} = \sum_m a_m u_m[/tex]
Since you already have the eigenvectors, it should be reasonably easy to find the coefficients am.

The possible results of a measurement of the z-component of angular momentum are given by the eigenvalues of the operator. The probability of each result is given by the coefficient of the corresponding eigenvector in the decomposition of the state. To put it in mathematical terms, with a system in the state [itex]x[/itex], the probability of obtaining [itex]m\hbar[/itex] as the z-angular momentum (for some particular choice of m) is
[tex]P_x(m) = \lvert a_m \rvert^2[/tex]

If you were using Dirac notation, that last formula could be written
[tex]P_x(m) = \lvert \langle u_m \vert x \rangle \rvert^2[/tex]
but the true meaning of that expression is a whole other can of worms which I'll refrain from getting into now
diazona
#3
Sep14-10, 07:56 PM
HW Helper
P: 2,155
Oh, and once you have the probability of each result, the expectation value is just a weighted average of the possible outcomes (the eigenvalues).
[tex]\langle L_z\rangle = \langle x\rvert L_z \lvert x\rangle = \sum_m P_x(m) m\hbar[/tex]


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