# Model for a Qubit system using the Hamiltonion Operator

• Lambda96
In summary, the conversation discusses a mathematical calculation involving eigenvalues and eigenvectors. The speaker is not sure if they have calculated it correctly and requests help. They mention using the usual formula for calculating eigenvalues and provide the resulting values. They also mention using Mathematica to calculate the normalized eigenvectors, but the other person suggests using LaTeX instead. The conversation then goes on to discuss the correct symbols to use and the importance of showing work. The speaker then confirms their calculations and provides the normalized eigenvectors. The other person explains the importance of using a substitution to make the algebra more transparent. The conversation ends with a discussion about verifying the eigenvectors using a calculation.
Lambda96
Homework Statement
Calculate the eigenvalues and eigenvectors of the Hamilton operator
Relevant Equations
none
Hi,

unfortunately, I am not sure if I have calculated the task a correctly.

I calculated the eigenvalues with the usual formula ##\vec{0}=(H-\lambda I) \psi## and got the following results

$$\lambda_1=E_1=-\sqrt{B^2+\nabla^2}$$
$$\lambda_2=E_2=\sqrt{B^2+\nabla^2}$$

I'm just not sure about the normalized eigenvectors, I got the following as eigenvectors

$$\vec{\psi}_1= \left(\begin{array}{c} \frac{B- \sqrt{B^2+\triangle^2}}{\triangle} \\ 1 \end{array}\right)$$
$$\vec{\psi}_2= \left(\begin{array}{c} \frac{B+ \sqrt{B^2+\triangle^2}}{\triangle} \\ 1 \end{array}\right)$$

For the normalization, I must divide the vectors by their norm. Since writing this vector via Latex is very time-consuming, I have calculated this with Mathematica, instead of the triangle, I have simply written an A. For the eigenvector ##\vec{\psi_1}## I got the following:

Since this eigenvector looks very messy, I am not sure if I have calculated the task correctly now.

First you need to be less confusing with your symbols. Use \Delta for ##\Delta## and not ##\nabla##, ##\triangle## or ##A##.
According to our rules, to receive help, you need to show your work. The derivation of the normalized eigenvectors is straightforward but I cannot guide you to it unless I see what what you did. Try to use LaTeX, but if you must use Mathematica, did you know that you can export its equations in LaTeX?

Your expression for ##|\psi_1\rangle## is not correct because it is not a constant times the unnormalized ##|\psi_1\rangle##. If you factor out the denominator, you are left with $$\begin{pmatrix} \frac{-B-\sqrt{B^2+\Delta ^2}}{\Delta} & 1 \end{pmatrix}$$while the unnormalized vector is
$$\begin{pmatrix} \frac{B-\sqrt{B^2+\Delta ^2}}{\Delta} & 1 \end{pmatrix}.$$I suggest that you replace the radical with a single symbol, e.g. ##R\equiv \sqrt{B^2+\Delta^2}.## It will make the algebra more transparent.

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Lambda96

Sorry, because of all the different symbols, unfortunately I still had in mind that it is the Nabla operator ##\nabla##. I had then changed all symbols by ##\triangle## and forgotten the change in the eigenvalues. But as you already said correctly, it should be ##\Delta##
kuruman said:
Try to use LaTeX, but if you must use Mathematica, did you know that you can export its equations in LaTeX?
I did not know that, thanks for the tip
I have now been able to confirm by the following calculation that these are the eigenvectors of A

$$H \vec{\psi}_1=E_1 \vec{\psi}_1$$
$$H \vec{\psi}_2=E_2 \vec{\psi}_2$$

The normalized eigenvectors are then as follows.

$$\hat{\psi}_1=\frac{1}{\sqrt{\biggl( \frac{(B-\sqrt{B^2+\Delta^2})}{\Delta} \biggr)^2+1}} \cdot \left(\begin{array}{c} \frac{B-\sqrt{B^2+\Delta^2}}{\Delta} \\ 1 \end{array}\right)$$

$$\hat{\psi}_2=\frac{1}{\sqrt{\biggl( \frac{(B+\sqrt{B^2+\Delta^2})}{\Delta} \biggr)^2+1}} \cdot \left(\begin{array}{c} \frac{B+\sqrt{B^2+\Delta^2}}{\Delta} \\ 1 \end{array}\right)$$

This is how I would do it.

The unnormalized eigenvector corresponding to ##E_1## is $$|\psi_1\rangle= \begin{pmatrix} \frac{B-R}{\Delta} \\ 1 \end{pmatrix}~~~R\equiv \sqrt{B^2+\Delta^2}.$$Let ##N## be the normalization constant by which you must multiply the eigenvector. Then \begin{align} 1= & N^2\langle \psi_1 | \psi_1\rangle = \left(\frac{B-R}{\Delta}\right)^2+1= \frac{B^2-2BR+R^2+\Delta^2}{\Delta^2}=\frac{2R^2-2BR}{\Delta^2 }=\frac{2R(R-B)}{\Delta^2} \nonumber \\ & \implies N=\frac{\Delta}{\sqrt{2R(R-B)}} \nonumber \end{align}and the normalized eigenvector is
$$|\psi_1\rangle=\frac{1}{\sqrt{2R(R-B)}} \begin{pmatrix} {B-R} \\ \Delta \end{pmatrix}.$$The normalized eigenvector corresponding to ##E_2## is obtained by swapping the column vector elements and introducing a relative minus sign. Obviously, the overall phase doesn't matter.
$$|\psi_2\rangle=\frac{1}{\sqrt{2R(R-B)}} \begin{pmatrix} -\Delta \\ B-R \end{pmatrix}.$$It is immediately clear that the above vectors are orthonormal. In your expression that is not obvious. That is what I meant when I said that the substitution ##R=\sqrt{B^2+\Delta^2}## in such problems makes the algebra transparent.

Lambda96
Thanks again kuruman for your help and for the tip with the calculation of the normalization constant ##N## .

I would have only one question, concerning the normalized eigenvector ##\hat{\psi}_2##.

I would have now calculated the normalization constant exactly as for ##\hat{\psi}_1## and got the following

$$N=\frac{\Delta}{\sqrt{2R(R+B)}}$$

I would then get the following as the normalized eigenvector.

$$\hat{\psi}_2=\frac{\Delta}{\sqrt{2R(R+B)}} \cdot \left(\begin{array}{c} B+R \\ 1 \end{array}\right)$$

Unfortunately, I don't quite understand why I also get the eigenvector when I swap the columns for the normalized eigenvector ##\hat{\psi}_1## and multiply by minus. I wanted to verify with the calculation ##H\hat{\psi}_2=E_2 \hat{\psi}_2## that it is an eigenvector of H and unfortunately I got the following.

$$H \cdot \left(\begin{array}{c} -\Delta \\ B-R \end{array}\right)=\frac{\Delta}{\sqrt{2R(R+B)}} \cdot \left(\begin{array}{c} -\Delta R\\ -\Delta^2-B(B-R) \end{array}\right)$$

The first line is correct, only with the second line I have problems.

Is it possible to determine the normalized eigenvectors of a 2x2 matrix with this trick, or only in certain cases?

First of all this $$\hat{\psi}_2=\frac{\Delta}{\sqrt{2R(R+B)}} \cdot \left(\begin{array}{c} B+R \\ 1 \end{array}\right)$$is dimensionally incorrect. The first element in the column vector has dimensions of energy while the second element is dimensionless. That cannot be. Also, I think you should have a term ##(R-B)## in the denominator, not ##(R+B).## Recheck your work.

Lambda96 said:
Is it possible to determine the normalized eigenvectors of a 2x2 matrix with this trick, or only in certain cases?
It is possible to use this trick with any 2×2 matrix. In the more general case where the eigenvectors have complex coefficients, let the normalized eigenvectors be
##|\psi_1\rangle=a_1|1\rangle +b_1|2\rangle~~~(|a_1|^2+|b_1|^2=1).##
##|\psi_2\rangle=a_2|1\rangle +b_2|2\rangle~~~(|a_2|^2+|b_2|^2=1).##
To ensure that they are orthogonal, you must have
$$0=\langle\psi_2|\psi_1\rangle= \begin{pmatrix} {a^*_2} & b^*_2 \end{pmatrix} \begin{pmatrix} {a_1} \\ b_1 \end{pmatrix}=a^*_2a_1+b^*_2b_1.$$Once you obtain ##a_1## and ##b_1## for the first eigenvector, you can immediately write ##a_2\rightarrow -b^*_1## and ##b_2\rightarrow a^*_1## because then $$a^*_2a_1+b^*_2b_1\rightarrow (-b^*_1)^*a_1+ (a^*_1)^*b_1=-b_1a_1+a_1b_1=0.$$Of course, if the coefficients are real as in this case, you don't have to worry about the complex conjugates, just swap elements and introduce a relative negative sign.

Lambda96
Thanks kuruman for your help and your explanation for finding the eigenvectors of a 2x2 matrix, for future tasks, this will simplify the calculation

## What is a Hamiltonian operator in the context of a qubit system?

A Hamiltonian operator in the context of a qubit system is a mathematical operator that describes the total energy of the system. It is used to determine the evolution of the qubit's state over time according to the Schrödinger equation. The Hamiltonian encapsulates all the interactions and energy contributions within the system.

## How does the Hamiltonian operator affect the state of a qubit?

The Hamiltonian operator affects the state of a qubit by governing its time evolution. When a Hamiltonian is applied to a qubit, it influences the probability amplitudes of the qubit's basis states, causing the qubit to evolve in a specific way over time. This is described by the time-dependent Schrödinger equation, where the Hamiltonian determines the rate and manner of change in the qubit's state.

## What is the significance of eigenvalues and eigenvectors of the Hamiltonian in a qubit system?

The eigenvalues and eigenvectors of the Hamiltonian in a qubit system are significant because they represent the possible energy levels and corresponding states of the system. The eigenvalues correspond to the measurable energy levels, while the eigenvectors (or eigenstates) are the specific states associated with these energy levels. These eigenstates form a basis in which the qubit's state can be expressed and analyzed.

## Can you provide an example of a simple Hamiltonian for a single qubit?

A simple example of a Hamiltonian for a single qubit is the Pauli-X operator, often represented as $$H = \frac{\hbar \omega}{2} \sigma_x$$, where $$\hbar$$ is the reduced Planck's constant, $$\omega$$ is the angular frequency, and $$\sigma_x$$ is the Pauli-X matrix. The Pauli-X matrix is given by:$\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$This Hamiltonian causes the qubit to oscillate between its basis states |0⟩ and |1⟩.

## How do you solve the Schrödinger equation for a qubit system with a given Hamiltonian?

To solve the Schrödinger equation for a qubit system with a given Hamiltonian, you typically follow these steps:1. Identify the Hamiltonian operator $$H$$ of the system.2. Write down the time-dependent Schrödinger equation: $$i\hbar \frac{\partial}{\partial t} |\psi(t)\rangle = H |\psi(t)\rangle$$.3. If the Hamiltonian is time-independent, solve the eigenvalue equation \( H |\phi

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