# Understanding the time evolution of a quantum state

I am trying to understand this equality below, but I can't seem to wrap my head around it. The Hamiltonian is defined as $\vec H=-\frac {\gamma B \hbar}{2}\sigma_x$, which gives the eigenvalues $\pm \frac {\gamma B \hbar}{2}$.

$$|\psi (t)\rangle = e^{-iHt/ \hbar} \chi_+ = \left(\cos\left(\frac{Ht}{\hbar}\right)-i\sin\left(\frac{Ht}{\hbar}\right)\right)\chi_+=$$

$$=\left(\cos\left(\frac{B\gamma t}{2}\right)I-i\sin\left(\frac{B\gamma t}{2}\right)\sigma_x\right)\chi_+$$

It's the last equality I'm stuck at. Any help would be greatly appreciated.

I've tried writing it all out as matrices, and this is what I get from the left part of the equality:

$$\left(\cos\left(\frac{Ht}{\hbar}\right)-i\sin\left(\frac{Ht}{\hbar}\right)\right)\begin{pmatrix} 1 \\ 0 \end{pmatrix}= \begin{pmatrix} \cos\left(\frac{Ht}{\hbar}\right)-i\sin\left(\frac{Ht}{\hbar}\right)\\0 \end{pmatrix}$$ The above should be equal to the right hand side of the equality, but this is what I get instead: $$\left( \cos\frac {B\gamma t}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} -i\sin\frac{B\gamma t}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right) \begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} \cos\left(\frac{B\gamma t}{2}\right) \\ -i\sin\left(\frac{B\gamma t}{2}\right) \end{pmatrix}$$

• If you write out your last as $2\times 2$ matrices and $\chi_+$ as a column vector it will come to you. – ZeroTheHero Oct 8 '17 at 13:21
• See edit, I have tried this but I'm doing something wrong, can't get them to look the same. – armara Oct 8 '17 at 13:35
• Your $e^{-itH/\hbar}=\cos(Ht/\hbar)-i\sin(Ht/\hbar)$ should be a matrix because $H$ is a matrix. – ZeroTheHero Oct 8 '17 at 13:47
• Thanks! I didn't understand how I was supposed to work with a matrix inside my cosine- and sinus-functions. – armara Oct 8 '17 at 14:06
• Btw regarding this one, physics.stackexchange.com/questions/362611/…, there's no need to delete posts marked as duplicates - they serve as useful waymarkers for future visitors. – Emilio Pisanty Oct 13 '17 at 13:56

## 3 Answers

So we define $f(M)$ for matrices $M$ and functions $f$ by appealing to their Taylor series: we find $f(x) = f(0) + f'(0)~x + \frac12 f''(0)~x^2 + \dots$ and then we extend this to matrices with $$f(M) = f(0) + f'(0)~M + \frac12 f''(0)~M^2 + \dots.$$This appears to be different from what you're doing, which (correct me if I'm wrong) is something more like $$f\left(\begin{bmatrix}M_{11}&M_{12}\\ M_{21}&M_{22}\end{bmatrix}\right)=\begin{bmatrix}f(M_{11})&f(M_{12})\\ f(M_{21})&f(M_{22})\end{bmatrix}.$$ While the above is wrong please do not beat yourself up too much about it: There is a way to make your intuition match the standard procedure, and it involves finding $C,C^{-1}$ such that $C^{-1}MC$ is a diagonal matrix $\operatorname{diag}(a,b,c\dots)$: once you have this, one can prove that the product of two diagonal matrices is diagonal, with the diagonal elements multiplied: $\operatorname{diag}(a, b, c\dots)~\operatorname{diag}(p, q, r\dots) = \operatorname{diag}(ap,~ bq,~ cr~\dots).$ If we plug this into the above "correct" definition we find that function applications distribute over diagonal entries in a diagonal matrix,$$f(\operatorname{diag}(a,b,c\dots)) = \operatorname{diag}(f(a),~f(b),~f(c)~\dots).$$

For the Pauli matrices we can somewhat sidestep this as $\sigma_i^2 = I$ and therefore the above Taylor expansion reduces to simply $$f(\alpha \sigma_i) = \frac{f(\alpha) + f(-\alpha)}2 ~ I + \frac{f(\alpha) - f(-\alpha)}2 ~ \sigma_i,$$ where the first term is readily recognized as the even part of $f$ and the second term is readily recognized as the odd part of $f$. The basic idea is that the even part of $f$ contains in its Taylor expansion only the even powers $x^{2n}$ while the odd part of $f$ contains only the odd powers $x^{2n+1}.$

• Thank you! This is what I needed, I didn't know how to work with matrices inside of a function (sinus/cosinus). – armara Oct 9 '17 at 7:36

That's ok. Your vector $$\left(\begin{array}{cc} \cos \omega t \\ -i\sin \omega t \end{array}\right) \tag{1}$$ is the result of time-evolving your initial state $(1,0)^T$. Since this initial state is not an eigenstate of your Hamiltonian, this initial state will evolve to a mixture of basis states, and (1) is just this. Eq.(1) just states that your initial state will evolve so that, after time $t$, it will be found in the state $\chi_+$ with probability $\cos(\omega t)^2$ and in the state $\chi_-$ with probability $\sin(\omega t)^2$. Note of course that the sum of these probabilities is $1$, as it should be.

What I needed to use was: For more information on the topic: https://en.wikipedia.org/wiki/Trigonometric_functions_of_matrices

• You might want to delete this... – ZeroTheHero Oct 8 '17 at 14:14

## protected by Qmechanic♦Oct 8 '17 at 14:22

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?