# 71GA

reputation
112
bio
website        ziga-lausegger.netau.net/…
location       Slovenia
age            27
member for     1 year, 5 months
seen           2 mins ago
profile views  174

I love to program and crosscompile baremetal C programs for ARM based microcontrollers, I love physics and i love writing science documents/books in LaTeX. It amazes me how physics is connecting all science and is helping mathematics to evolve. In order for science profession to comunicate on a high level i advise everyone to use Linux, LaTeX and a good vector imaging program like Inkscape.

|            |                | bio      | visits                     |            |                  |
|------------|----------------|----------|----------------------------|------------|------------------|
|            | 366 reputation | website  | ziga-lausegger.netau.net/… | member for | 1 year, 5 months |
| 112 badges | location       | Slovenia | seen                       | 2 mins ago |                  |

# 290 Actions

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| 4h    | comment  | Vector $\vec{z}$ and its conjugate transpose $\overline{\vec{v}^\top}$ - is it the same as $\left|z\right\rangle$ and $\left\langle z \right|$Interesting but i don't understand quite that much yet. I hope i will in the future.                                                                                                                                                                                                                                                                                                                                                                                                                                                       |
| 4h    | accepted | Vector $\vec{z}$ and its conjugate transpose $\overline{\vec{v}^\top}$ - is it the same as $\left|z\right\rangle$ and $\left\langle z \right|$                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           |
| 4h    | comment  | Vector $\vec{z}$ and its conjugate transpose $\overline{\vec{v}^\top}$ - is it the same as $\left|z\right\rangle$ and $\left\langle z \right|$Dirac notation is now somehow clearer to me :)                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             |
| 4h    | comment  | Vector $\vec{z}$ and its conjugate transpose $\overline{\vec{v}^\top}$ - is it the same as $\left|z\right\rangle$ and $\left\langle z \right|$If i look closer now i can see that $\left\langle A|B \right\rangle$ is an inner product between $\left|B\right\rangle$ and $\left|A\right\rangle$. This is correct right? Can I allso say that this is a matrix multiplication of an $\left|A\right\rangle^\dagger$ and $\left|B\right\rangle$?                                                                                                                                                                                                                                           |
| 5h    | comment  | Vector $\vec{z}$ and its conjugate transpose $\overline{\vec{v}^\top}$ - is it the same as $\left|z\right\rangle$ and $\left\langle z \right|$So my assumptions are correct :D TY!                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                       |
| 6h    | comment  | Vector $\vec{z}$ and its conjugate transpose $\overline{\vec{v}^\top}$ - is it the same as $\left|z\right\rangle$ and $\left\langle z \right|$But they are all connected and i like it better like this.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 |
| 7h    | asked    | Vector $\vec{z}$ and its conjugate transpose $\overline{\vec{v}^\top}$ - is it the same as $\left|z\right\rangle$ and $\left\langle z \right|$                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           |
| May12 | comment  | Some Dirac notation explanationsThank you i think i now understand a bit more.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           |
| May12 | asked    | Some Dirac notation explanations                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         |
| May11 | accepted | How do we know that $\psi$ is the eigenfunction of an operator $\hat{H}$ with eigenvalue $W$?                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            |
| May11 | comment  | How do we know that $\psi$ is the eigenfunction of an operator $\hat{H}$ with eigenvalue $W$?This Dirac notation is confusing for a starters ... I tried reading Zetilli and got lost ... there is so much of this stuff / rules ...                                                                                                                                                                                                                                                                                                                                                                                                                                                     |
| May10 | comment  | How do we know that $\psi$ is the eigenfunction of an operator $\hat{H}$ with eigenvalue $W$?Thank you for this explaination. It was brief and provided lots of good info. There is only one more thing. I don't quite understand this equation: $\langle O \rangle = \langle \Psi | O | \Psi \rangle$. Is this a scalar product with itself? And then an operator acts on this scalar product? I know that if we use a $\dagger$ on a ket we get a bra, so it must hold that: $\langle O\rangle = \langle \psi |O| \psi \rangle = |\psi\rangle^\dagger O|\psi\rangle$... But where is the integral? Shouldnt it be: $\langle O\rangle = \int |\psi\rangle^\dagger O|\psi\rangle d x$ ? |
| May10 | comment  | How do we know that $\psi$ is the eigenfunction of an operator $\hat{H}$ with eigenvalue $W$?I dont know if you understood my question right. How do we know from this $\langle W \rangle = \int \limits_{-\infty}^{\infty} \overline{\Psi}\, \left(- \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p\right) \Psi \, d x$ or this $\hat{H} = - \frac{\hbar^2}{2m} \frac{d^2}{d \, x^2} + W_p$ that we have an eigenfunctiuion and eigenvalue.                                                                                                                                                                                                                                             |
| May10 | asked    | How do we know that $\psi$ is the eigenfunction of an operator $\hat{H}$ with eigenvalue $W$?                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            |
| May9  | comment  | QM formalism is one big confusion - lack of geometrical explaination with imagesThank you. It seems i complicate too much.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               |
| May8  | comment  | QM formalism is one big confusion - lack of geometrical explaination with imagesAnd one more questions... It is clear to me that if bra's are column vectors then kets are row vectors. But i don't know the physical meaning of bra's (do they even have one?). I know that ket's are QM states.                                                                                                                                                                                                                                                                                                                                                                                        |
| May8  | comment  | QM formalism is one big confusion - lack of geometrical explaination with imagesSo if i understood right we choose $\mathbb{R}^{n}$ as a space for probabilities, BUT in QM we have amplitudes and an amplitude is a square root of a probability. So i need a complex space $\mathbb{C}^n$ because of the square root which leads to complex numbers?                                                                                                                                                                                                                                                                                                                                   |
| May7  | comment  | Energy eigenvalues of a Q.H.Oscillator with $[\hat{H},\hat{a}] = -\hbar \omega \hat{a}$ and $[\hat{H},\hat{a}^\dagger] = \hbar \omega \hat{a}^\dagger$Thank you! the most helpfull was the fact that $\hat{H} \neq W_n$ but $\hat{H} \psi_n = W_n \psi_n$. It looks a bit WIERD though...                                                                                                                                                                                                                                                                                                                                                                                                |
| May7  | accepted | Energy eigenvalues of a Q.H.Oscillator with $[\hat{H},\hat{a}] = -\hbar \omega \hat{a}$ and $[\hat{H},\hat{a}^\dagger] = \hbar \omega \hat{a}^\dagger$                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                   |
| May7  | accepted | QM formalism is one big confusion - lack of geometrical explaination with images                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         |
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