• Support PF! Buy your school textbooks, materials and every day products Here!

Quantum Mechanics: zero eigenvector

  • Thread starter Telemachus
  • Start date
  • #1
832
30
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:

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,200
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}##
Your eigenvalues aren't correct. They should be ##E_0 \pm \sqrt{2}a## and ##E_0##.

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.
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
832
30
Thank you Vela. I've found the mistake. I'll be back :D
 
  • #4
832
30
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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,200
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
832
30
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:
  • #7
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,200
Yes, you're right. It looks like sometimes, you're getting a total probability exceeding 1.
 
  • #8
832
30
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:
  • #9
832
30
Damn, I realized I didn't find the mistake as I thought haha sorry for that.
 
  • #10
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,200
I found it while transcribing to latex. Thanks.

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

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}}##
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.

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{\sqrt{2}}{8}e^{-i\frac{\sqrt{2}at}{\hbar}}+\frac{\sqrt{2}}{8}e^{-i\frac{\sqrt{2}at}{\hbar}}-\frac{\sqrt{2}}{8}e^{i\frac{\sqrt{2}at}{\hbar}}-\frac{1}{8}e^{-i\frac{\sqrt{2}at}{\hbar}}##

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


EDIT: Ok, the mistake was in P(C), I missed a half somewhere:

##P(C)=\frac{1}{2}-\frac{1}{2}\cos(\frac{2\sqrt{2}a}{\hbar}t)##
 
Last edited:
  • #11
832
30
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:
  • #12
832
30
This simplifies to ##e^{-i E_0 t/\hbar} \cos^2\left(\frac{a t}{\sqrt{2}\hbar}\right)##.
This was right, I've checked it.

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.
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:
  • #13
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,200
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})##
You made a mistake in the last step. You should have
$$\frac{1}{2}e^{-i E_0 t/\hbar} \left( -1 + \cos \frac{\sqrt{2}at}{\hbar} \right) = -e^{-i E_0 t/\hbar} \left( \frac{1 - \cos \frac{\sqrt{2}at}{\hbar}}{2}\right) = -e^{-i E_0 t/\hbar}\sin^2 \frac{at}{\sqrt{2}\hbar}$$

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?
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.
You're correct about the frequencies.

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
832
30
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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,200
You're making it too complicated. Look at where are A, B, and C relative to B.
 
  • #16
832
30
Ok, I think I could say they are at a distance -d, 0 and d from B. Is that what you mean?
 
  • #17
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,200
Yup.
 

Related Threads on Quantum Mechanics: zero eigenvector

Replies
4
Views
555
  • Last Post
Replies
9
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
7
Views
467
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
18
Views
1K
Replies
1
Views
972
Top