# Quantum Mechanics: zero eigenvector

1. Sep 19, 2013

### Telemachus

Well, I know this have no sense. But I was trying to solve a problem on Cohen Tannoudji. The problem is in chapter IV, complement $J_{IV}$, exercise 8. It says:

Consider an electron of a linear triatomic molecule formed by three equidistant atoms. We use $\left | {\phi_A} \right >, \left | {\phi_B} \right >, \left | {\phi_C} \right >$ to denotate three orthonormal states of this electron, corresponding respectiveley to three wave functions localized about the nuclei of atoms A, B, C. We shall confine ourselves to the subspace of the state space spanned by $\left | {\phi_A} \right >, \left | {\phi_B} \right >, \left | {\phi_C} \right >$.

When we neglect the possibility of the electron jumping from one nucleus to another, its energy is described by the Hamiltonian $H_0$ whose egienstates are the three states $\left | {\phi_A} \right >, \left | {\phi_B} \right >, \left | {\phi_C} \right >$ with the same eigenvalue $E_0$. The coupling between the states $\left | {\phi_A} \right >, \left | {\phi_B} \right >, \left | {\phi_C} \right >$ is described by an additional Hamiltonian defined by:

$W\left | {\phi_A} \right >=-a \left | {\phi_B} \right > \\ W\left | {\phi_B} \right >=-a \left | {\phi_A} \right >-a \left | {\phi_C} \right > \\ W\left | {\phi_C} \right >=-a \left | {\phi_B} \right >$

Where $a$ is a real positive constant.

1. Calculate the energies and stationary states of the Hamiltonian $H=H_0+W$.

2. The electron at time t=0 is in the state $\left | {\phi_A} \right >$. Discuss qualitatively the localization of the electron at subsequent times t. Are there any values of t for which it is perfectly localized about atom A,B, or C?

3. Let $D$ be the observable whose eigenstates are $\left | {\phi_A} \right >, \left | {\phi_B} \right >, \left | {\phi_C} \right >$ with respective eigenvalues $-d,0,d$. $D$ is measured at time t; what values can be found, and with what probabilities?

4. When the initial state of the electron is arbitrary, what are the Bohr frequencies that can appear in the evolution of $\left < D \right >$? Give a physical interpretation of $D$. What are the frequencies of the electromagnetic waves that can be absorbed or emmitted by the molecule?

I transcribed the whole exercise, but I got stuck in part 1. I evaluated the matrix elements for W.

I've got the Hamiltonian:

$H= \begin{pmatrix} E_0 & -a & 0 \\ -a & E_0 & -a \\ 0 & -a & E_0 \\ \end{pmatrix}$

When I diagnoalize it, I get the three eigenvalues: $E_1=E_0,E_2=E_0-\sqrt{E_0^2+2a^2},E_3=E_0+\sqrt{E_0^2+2a^2}$

Then, I've tried to find the eigenstates,
For $E_1$:

$\left | {E_1} \right >= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0\\ -1\\ \end{pmatrix}$

But then, when I try to get $\left | E_1 \right >$ I get intro trouble. I have the system of equations:

$(E_0-E)\beta-a\gamma=0 \\ -a\beta+(E_0-E)\gamma-a\eta=0 \\ -a\gamma+(E_0-E)\eta=0$

So I get $\gamma=0,\beta=0,\eta=0$. I get the zero vector as the eigenvector. And thats absurd, because any eigenvalue could be an eigenvalue for the zero eigenvector. I don't know if I did something wrong, or if I have to interpret this result somehow.

Thanks in advance for your help. I have transcribed the whole exercise because, if I get some help and I can go on with it, I would like to discuss the other results that I could find.

Last edited: Sep 19, 2013
2. Sep 19, 2013

### vela

Staff Emeritus
Your eigenvalues aren't correct. They should be $E_0 \pm \sqrt{2}a$ and $E_0$.

My guess is that since you're not substituting in an actual eigenvalue, you're not getting a dependent system of equations, which in turn leads to only the trivial solution.

3. Sep 19, 2013

### Telemachus

Thank you Vela. I've found the mistake. I'll be back :D

4. Sep 19, 2013

### Telemachus

Ok. Now my eigenstates are:
$\left | {E_1} \right >= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0\\ -1\\ \end{pmatrix} \: \left | {E_2} \right >= \begin{pmatrix} 1/2 \\ 1/\sqrt{2}\\ 1/2\\ \end{pmatrix} \: \left | {E_3} \right >= \begin{pmatrix} -1/2 \\ 1/\sqrt{2}\\ -1/2\\ \end{pmatrix}$

Then, for point 2:

$\left | {\psi (0)} \right >=\left | {\phi_A} \right >= \begin{pmatrix} 1 \\ 0\\ 0\\ \end{pmatrix}$

So: $\left | {\psi (t)} \right >=\frac{1}{\sqrt{2}}e^{-i \frac{E_0}{\hbar}t}\left | {E_1} \right >+\frac{1}{2}e^{-i \frac{E_0-\sqrt{2}a}{\hbar}t}\left | {E_2} \right >-\frac{1}{2}e^{-i \frac{E_0+\sqrt{2}a}{\hbar}t}\left | {E_3} \right >$

To answer 2. I thought of computing the probability of finding the system in states $\left | {\phi_A} \right >, \left | {\phi_B} \right >, \left | {\phi_C} \right >$. But I'm not sure if that's what the problem asks. What you think Vela?

Thank you again.

5. Sep 19, 2013

### vela

Staff Emeritus
Yes, that's what the problem is asking. Since you're interested in where the electron is, it might make sense to express the state in terms of $\lvert \phi_A \rangle$, etc., rather than in terms of the energy eigenstates.

6. Sep 20, 2013

### Telemachus

Hello again Vela.

I did as you said. Then I found that the probability of A:

$P(A)= \frac{3}{8} + \frac{1}{2} \cos(\frac{\sqrt{2}a}{\hbar}t) + \frac{1}{8} \cos(\frac{2 \sqrt{2} a}{\hbar}t)$

From here it was clear to me that for certain times $t=\frac{n\pi\hbar}{2\sqrt{2}a},n=0,2,4...$ the probability of finding the electron perfectly localized around atom A would be 1.

But then I calculated the probability for C, and I got:

$P(C)=\frac{1}{2}-\frac{1}{4}\cos(\frac{2\sqrt{2}a}{\hbar}t)$
And the problem is that for those times, I have 1/4 of probability of finding the atom localized about C. So, I did something wrong, right? the probability for those times of finding the electron around C or B should be zero.

Last edited: Sep 20, 2013
7. Sep 20, 2013

### vela

Staff Emeritus
Yes, you're right. It looks like sometimes, you're getting a total probability exceeding 1.

8. Sep 20, 2013

### Telemachus

Ok, this is what I did:

First of all I expressed the state in terms of the eigeinstates for the Hamiltonian $H_0$:

$\left | {\psi (t)} \right > = \frac{1}{2} e^{-i\frac{E_0t}{\hbar} } \left ( \left | {\phi_A} \right > - \left | {\phi_C} \right > \right ) + \frac{1}{4} e^{-i \frac{ (E_0-\sqrt{a} )t}{\hbar}} \left ( \left | {\phi_A} \right > +\sqrt{2}\left | {\phi_B} \right > + \left | {\phi_C} \right > \right ) +\frac{1}{4} e^{-i\frac{(E_0+\sqrt{a})t}{\hbar}} \left ( \left | {\phi_A} \right > - \sqrt{2} \left | {\phi_B} \right >+ \left | {\phi_C} \right > \right )$

As a check I did $\left | {\psi (0)} \right >= \left | {\phi_A} \right >$

Then:

$\left < {\phi_A} \right | \left | {\psi (t)} \right >=\frac{1}{2}e^{-i\frac{E_0t}{\hbar}}+ \frac{1}{4}e^{-i\frac{(E_0-\sqrt{a})t}{\hbar}}+\frac{1}{4}e^{-i\frac{(E_0+\sqrt{a})t}{\hbar}}$

Then taking the complex conjugate and making the product I found $P(A)$

Similarly:

$\left < {\phi_C} \right | \left | {\psi (t)} \right >=-\frac{1}{2}e^{-i\frac{E_0t}{\hbar}}+ \frac{\sqrt{2}}{4}e^{-i\frac{(E_0-\sqrt{a})t}{\hbar}}-\frac{\sqrt{2}}{4}e^{-i\frac{(E_0+\sqrt{a})t}{\hbar}}$

And then I found $P(C)$ taking the complex conjugate:

$P(C)=| \left < {\phi_C} \right | \left | {\psi (t)} \right > |^2= \left ( -\frac{1}{2}e^{-i\frac{E_0t}{\hbar}}+ \frac{\sqrt{2}}{4}e^{-i\frac{(E_0-\sqrt{a})t}{\hbar}}-\frac{\sqrt{2}}{4}e^{-i\frac{(E_0+\sqrt{a})t}{\hbar}} \right ) \left ( -\frac{1}{2}e^{i\frac{E_0t}{\hbar}}+ \frac{\sqrt{2}}{4}e^{i\frac{(E_0-\sqrt{a})t}{\hbar}}-\frac{\sqrt{2}}{4}e^{i\frac{(E_0+\sqrt{a})t}{\hbar}} \right ) \\ = \frac{1}{4}+\frac{1}{8}+\frac{1}{8}- \frac{1}{8}e^{i\frac{2\sqrt{2}at}{\hbar}}-\frac{1}{8}e^{i\frac{2\sqrt{2}at}{\hbar}} \\ =\frac{1}{2}-\frac{1}{4}\cos(\frac{2\sqrt{2}a}{\hbar}t)$

Well, thank you very much vela. Just let me know if you see any evident mistake, because I couldn't find it.

Last edited: Sep 20, 2013
9. Sep 20, 2013

### Telemachus

Damn, I realized I didn't find the mistake as I thought haha sorry for that.

10. Sep 20, 2013

### vela

Staff Emeritus
This simplifies to $e^{-i E_0 t/\hbar} \cos^2\left(\frac{a t}{\sqrt{2}\hbar}\right)$.

This should simplify to $-e^{-iE_0t/\hbar}\sin^2\left(\frac{a t}{\sqrt{2}\hbar}\right).$ I didn't get factors of $\sqrt{2}$ in the coefficients of the second and third term, and the sign of the last term was positive.

Last edited: Sep 20, 2013
11. Sep 20, 2013

### Telemachus

There is a mistake in there, in the exponents, the energies are $E_0±\sqrt{2}a$ it was a typo. Thank you vela.

EDIT: I see you realized of that mistake. Thank you again.

Last edited: Sep 20, 2013
12. Sep 21, 2013

### Telemachus

This was right, I've checked it.

But I think this isn't right. You're right about the factors, I did a mistake somewhere, and interchanged the coefficients in the state psi between one phi B and one phi C.

This is what I've obtained:
$\left < {\phi_C} \right | \left | {\psi (t)} \right >=-\frac{1}{2} e^{-i\frac{E_0t}{\hbar}} + \frac{1}{4} e^{-i\frac{(E_0-\sqrt{2}a)t}{\hbar}} + \frac{1}{4}e^{-i\frac{(E_0+\sqrt{2}a)t}{\hbar}} \\ =\frac{1}{2}e^{-i\frac{E_0t}{\hbar}} \left ( -1 +\frac{1}{2}e^{-i\frac{\sqrt{2}at}{\hbar}}+\frac{1}{2}e^{i\frac{\sqrt{2}at}{\hbar}} \right )= \frac{1}{2}e^{-i\frac{E_0t}{\hbar}} \left ( -1 + \cos ( \frac{\sqrt{2}at}{\hbar}) \right )= -\frac{1}{2}e^{-i\frac{E_0t}{\hbar}}\sin^2 ( \frac{\sqrt{2}at}{\hbar})$

For point 3, I thought of:
$P(-d)=P(A),P(0)=P(B),P(d)=P(C)$

As the observables $D$ and $H_0$ has the same eigenvectors, the decomposition of psi in their eigensteits will have the same coefficients, adn then the probabilities would be equal. Right?

And in point 4. I thought that the Bohr frequencies will be given just by the energy difference between the eigenstates of the Hamiltonian divided by h. And that would be the frequencies that can be emitted or absorbed by the molecule. I think that is right, but I don't know the physical interpretation of D.

Last edited: Sep 21, 2013
13. Sep 21, 2013

### vela

Staff Emeritus
You made a mistake in the last step. You should have
$$\frac{1}{2}e^{-i E_0 t/\hbar} \left( -1 + \cos \frac{\sqrt{2}at}{\hbar} \right) = -e^{-i E_0 t/\hbar} \left( \frac{1 - \cos \frac{\sqrt{2}at}{\hbar}}{2}\right) = -e^{-i E_0 t/\hbar}\sin^2 \frac{at}{\sqrt{2}\hbar}$$

Right.

Look at the picture of the linear molecule given in the problem. Atom A corresponds to -d; atom B, to 0; and atom C, to +d. What do you think the letter "d" might represent?

14. Sep 22, 2013

### Telemachus

I thought of something like polarization (dipole perhaps?), because it's related to the electron localization and it's a measurable quantity, but I'm not sure.

15. Sep 22, 2013

### vela

Staff Emeritus
You're making it too complicated. Look at where are A, B, and C relative to B.

16. Sep 22, 2013

### Telemachus

Ok, I think I could say they are at a distance -d, 0 and d from B. Is that what you mean?

17. Sep 22, 2013

### vela

Staff Emeritus
Yup.