# Seemingly simple quantum mechanics/linear algebra problem

1. Oct 22, 2009

### Void123

1. The problem statement, all variables and given/known data

For some reason, I am making a trivial mistake somewhere. I just need to find the eigenvalues and normalized eigenvectors of the following matrix:

H = (1/sqrt(2))*Matrix(Row 1 Row 2 Row 3)

Row 1 = [0 -i 0]
Row 2 = [i 3 3]
Row 3 = [0 3 0]

(Sorry, I don't know the proper latex code for a matrix)

If someone can work this out and just provide the solutions so I can compare, I would really appreciate it.

2. Relevant equations

....

3. The attempt at a solution

I tried.

2. Oct 22, 2009

### jdwood983

Use the normal eigenvalue equation:

$$\det(H-\lambda\mathbb{I})=0$$

Solve for $$\lambda$$ (you should have 3 values since you have a $$3\times3$$ matrix).

For the eigenvectors,

$$\left(\begin{array}{ccc}0&-i&0 \\ i&3&3 \\ 0&3&0\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)=\lambda\left(\begin{array}{c}x\\y\\z\end{array}\right)$$

where you have to do this as many times as you have $$\lambda$$ values. That should help you a lot

PS: to write matrices in LaTeX, use the command \begin{array} & \end{array}. The following produced the matrix above:

\left(\begin{array}{ccc}0&-i&0 \\ i&3&3 \\ 0&3&0\end{array}\right)

3. Oct 22, 2009

### gabbagabbahey

Click on the $\LaTeX$ image below to see the code that generated it

$$H=\frac{1}{\sqrt{2}}\begin{pmatrix}0 & -i & 0 \\ i & 3 & 3 \\ 0 & 3 & 0 \end{pmatrix}$$

You've been here long enough to know that isn't what we do here.

Then show us!

4. Oct 22, 2009

### jdwood983

I forgot the $$\frac{1}{\sqrt{2}}$$ constant in front of my matrix for the eigenvectors, it should read:

$$\frac{1}{\sqrt{2}} \left(\begin{array}{ccc}0&-i&0 \\ i&3&3 \\ 0&3&0\end{array}\right)\left(\begin{array}{c}x\\y\ \z\end{array}\right)=\lambda\left(\begin{array}{c} x\\y\\z\end{array}\right)$$

5. Oct 22, 2009

### Void123

Thanks guys.

For my eigenvalues, I got $$\lambda = 0, -2, 5$$

For the last two eigenvalues, when I set up the linear equations to solve I get something of the sort $$cy = cy$$ for one of them. What does that mean for $$y$$? When the authors of my textbook do an example like this and run into the same situation, they postulate that $$y = 1$$. But how is that when the variable cancels out on both sides?

Thanks again.

6. Oct 22, 2009

### gabbagabbahey

Try it again without forgetting the factor of $1/\sqrt{2}$....

7. Oct 23, 2009

### jdwood983

Doesn't that answer your question? If the authors are allowed to set a canceled variable to be 1, doesn't that mean you can too? Usually you let the canceled variable be 1 for simplicity, but you can make it a billion if you really wanted, it'll just make your later calculations a bit more difficult.

But as Gabba suggested, you should add in that $$1/\sqrt{2}$$ factor to find your eigenvalues before you turn in your assignment.