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: $$ \left( \cos\frac {B\gamma t}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} -i\sin\frac{B\gamma t}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right) \begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} \cos\left(\frac{B\gamma t}{2}\right) \\ -i\sin\left(\frac{B\gamma t}{2}\right) \end{pmatrix}$$

  • $\begingroup$ If you write out your last as $2\times 2$ matrices and $\chi_+$ as a column vector it will come to you. $\endgroup$ – ZeroTheHero Oct 8 '17 at 13:21
  • $\begingroup$ See edit, I have tried this but I'm doing something wrong, can't get them to look the same. $\endgroup$ – armara Oct 8 '17 at 13:35
  • $\begingroup$ Your $e^{-itH/\hbar}=\cos(Ht/\hbar)-i\sin(Ht/\hbar)$ should be a matrix because $H$ is a matrix. $\endgroup$ – ZeroTheHero Oct 8 '17 at 13:47
  • $\begingroup$ Thanks! I didn't understand how I was supposed to work with a matrix inside my cosine- and sinus-functions. $\endgroup$ – armara Oct 8 '17 at 14:06
  • $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Oct 13 '17 at 13:56

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 $x^{2n+1}.$

  • $\begingroup$ Thank you! This is what I needed, I didn't know how to work with matrices inside of a function (sinus/cosinus). $\endgroup$ – 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:

enter image description here

For more information on the topic: https://en.wikipedia.org/wiki/Trigonometric_functions_of_matrices

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

protected by Qmechanic Oct 8 '17 at 14:22

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