# I How to change the Hamiltonian in a change of basis

1. Jul 28, 2016

### IanBerkman

Dear all,

The Hamiltonian for a particle in a magnetic field can be written as
$$\hat{H} = \frac{1}{2}g\mu_B\textbf{B}\cdot\boldsymbol\sigma$$
where $\boldsymbol\sigma$ are the Pauli matrices.

This Hamiltonian is written in the basis of the eigenstates of $\sigma_z$, but how is it written/transformed when it is in the basis of the eigenstates of $\sigma_x$?

2. Jul 28, 2016

### hilbert2

If $U$ is a unitary operator that transforms a vector $\left|\psi\right>$ written in the basis of $\sigma_z$ eigenstates to a form $U\left|\psi\right>=\left|\psi'\right>$ where it's in the basis of $\sigma_x$ eigenstates, then the operator $H$ must change to an operator $H'$ such that its expectation value is the same in the new basis: $\left<\psi'\right|H'\left|\psi'\right> = \left<\psi\right|H\left|\psi\right>$. This, of course can be accomplished by setting $H'=UHU^{-1}$, where $U^{-1}$ is the inverse of the unitary operator $U$. If you're comfortable with linear algebra, you can think of the state vectors as column vectors that have $N$ components and operators as $N\times N$ matrices, where $N$ can be very large or infinite. However, it is also possible to play with abstract vectors and operators without reference to any particular basis.

EDIT: also note that the bra vector corresponding to $\left|\psi'\right>$ is $\left<\psi'\right|=\left<\psi\right|U^{-1}$.

3. Jul 29, 2016

### IanBerkman

Thanks, that made it clear and I should be able to solve it!
One question: is the bra vector not given by $\langle \psi'| = \langle \psi| U^\dagger$ and for a matrix with only real components this becomes $U^T$?

4. Jul 29, 2016

### vanhees71

Sure, if $U$ is unitary you have (by definition) $U^{-1}=U^{\dagger}$.

5. Jul 29, 2016

### IanBerkman

All right, thanks. I found it more convenient to work with $U^\dagger$ than with $U^{-1}$.

6. Jul 29, 2016

### strangerep

Huh?

The expression $\textbf{B}\cdot\boldsymbol\sigma$ is rotation-invariant. In what sense is it "written in the basis of the eigenstates of $\sigma_z$"?

What am I missing?

7. Jul 30, 2016

### IanBerkman

Any vector working on the Hamiltonian will result in another 2-dimensional vector where the first component responds to the spin-up $(1, 0)^T$ and the second to the spin-down state $(0,1)^T$. Since the basis spanned by the eigenvectors of $\sigma_z$ is $$\left\{\begin{pmatrix} 1\\0 \end{pmatrix},\begin{pmatrix} 0\\1 \end{pmatrix} \right\}$$ any vector working on the Hamiltonian will be immediately represented in this basis.

Please correct me if I am wrong.

8. Jul 30, 2016

### mike_brv

I think that the form of the hamiltonian should be invaraint to choise of basis , meaning the hamiltonian will look the same no matter in what basis you work with. the only thing that will change is the representation of the pauli matrices according to the choise of basis.

9. Jul 30, 2016

### PeroK

In general, you should consider:

$\textbf{B}\cdot\ \textbf{S} = B_xS_x + B_yS_y + B_zS_z$

This operator is independent of your choice of basis for the spinors. If, however, you choose as the basis the eigenstates of $S_z$, then $\textbf{S} = \frac{\hbar}{2} \boldsymbol\sigma$

If you choose any other basis for your spinors, then you simply express $\textbf{S}$ in terms of the spin operators for this basis. For example, if you choose the eigenstates of $S_x$, then $\textbf{S}$ will be a different permutation of the Pauli matrices.

10. Jul 30, 2016

### PeroK

This may be a notational thing. I took $\boldsymbol\sigma$ to be the spin expressed in the specific form for eigenstates of $S_z$ and $\textbf{S}$ would be the general basis-independent notation for the spin operator.

11. Jul 30, 2016

### IanBerkman

Hm, I am stuck with another Hamiltonian at this moment. It is somewhat similar as the problem above so I will put it here (I could also make a new thread but I have been making too many lately )

I want to prove the following:
$$\langle\textbf{r}_1|\frac{1}{2m}\hat{p}^2+V(\hat{r})|\textbf{r}_2\rangle = \left\{-\frac{\hbar^2}{2m}\nabla_{\textbf{r}_1}^2+V(\textbf{r}_1)\right\}\delta(\textbf{r}_1-\textbf{r}_2)$$

I get the same expression, however, the operators on the RHS are still operators working on $\textbf{r}$ in my case.
$$-\frac{\hbar^2}{2m}\nabla^2\langle \textbf{r}_1|\textbf{r}_2\rangle+V(\hat{r})\langle\textbf{r}_1|\textbf{r}_2\rangle = \left\{-\frac{\hbar^2}{2m}\nabla^2+V(\hat{r})\right\}\delta(\textbf{r}_1-\textbf{r}_2)$$

12. Jul 30, 2016

### strangerep

Oh, I see -- thanks. There seems to be a partial language barrier, and the OP is using some incorrect terminology. I guess he means the spinor representation where $\sigma_z = diag(1,-1)$. In that case, the answer given in post #2 should hopefully be enough to find the right $U$.

13. Jul 31, 2016

### IanBerkman

I forgot to mention my notation. I used $\boldsymbol\sigma=(\sigma_x,\sigma_y,\sigma_z)$. Is this notation wrong?

Last edited: Jul 31, 2016
14. Jul 31, 2016

### blue_leaf77

Use the relation $\hat{r}|\mathbf r_1\rangle = \mathbf r_1|\mathbf r_1\rangle$ and the fact that $V(\hat r)$ is Hermitian.

15. Jul 31, 2016

### IanBerkman

All right, and is the following true?
$$\nabla|\textbf{r}_1\rangle=\nabla_{\textbf{r}_1}|\textbf{r}_1\rangle$$

16. Jul 31, 2016

### blue_leaf77

Not quite true, you cannot differentiate a vector. In order to deal with the kinetic energy part, you need to know what $\langle \mathbf r| \hat p |\psi\rangle$ is equal to. What you are looking for can be found by starting from the definition of the displacement operator $\exp(-i\hat p \cdot \mathbf R/\hbar )$ where $\mathbf R$ is a space vector (i.e. not an operator). When acting on a position ket, this operator displaces the ket it acts on by the amount equal to $\mathbf R$,
$$\exp(-i\hat p \cdot \mathbf R/\hbar ) |\mathbf r \rangle = |\mathbf r+\mathbf R \rangle$$
It follows immediately that, for an arbitrary state $|\psi\rangle$,
$$\langle \mathbf r |\exp(i\hat p \cdot \mathbf R/\hbar )|\psi\rangle = \langle \mathbf r|\psi\rangle + i \langle \mathbf r|\hat p|\psi\rangle \cdot \mathbf R/\hbar - \frac{1}{\hbar^2} \langle \mathbf r|(\hat p \cdot \mathbf R)^2|\psi\rangle + \ldots \\ = \langle \mathbf r + \mathbf R|\psi\rangle$$
Then comparing the above series with the Taylor expansion of $\langle \mathbf r + \mathbf R|\psi\rangle = \psi(\mathbf r + \mathbf R)$, you should see that
$$\langle \mathbf r| \hat p |\psi\rangle = -i\hbar \nabla_r \langle \mathbf r |\psi\rangle$$

17. Aug 1, 2016

### IanBerkman

All right, thank you. You guys are great with helping!