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| bio | website | ziga-lausegger.netau.net/… |
|---|---|---|
| location | Slovenia | |
| age | 27 | |
| visits | member for | 1 year, 5 months |
| seen | 2 hours ago | |
| stats | profile views | 177 |
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.
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2h |
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Eigenvalue $a_n$ I understand now, how to get an eigenvalue like $a_n = \langle\psi_n|\psi(t)\rangle$. What i was missing was the fact that eigenvectors are normalized. |
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2h |
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Eigenvalue $a_n$ About your anwser on Q1: But what if i would write $|\psi\rangle$ on the both sides instead of $|\psi\rangle$? Which one is correct? About your anwser on Q2: I can imagine at most 3 eigenvectors being orthogonal to each other and form a basis. All the rest i cannot visualize being orthogonal, but how is it possible? Is this just an abstraction and i shouldnt bother so much with it? I vizualize inner product is an analog to scalar product which is a projection of 1st vector's norm to the other vec. multiplied by a 2nd vector's norm. And vice versa - where order matters at inner pr. |
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5h |
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Some Dirac notation explanations So the general importance here is that matrix multiplication has to be defined and thie is why it is important if we use operator from the left or the right :) |
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1d |
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Some Dirac notation unclarities Have you read Zetilli's book? Is the chapter "Postulates of QM" the thing i need to read? It sure looks nice... |
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1d |
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Some Dirac notation unclarities Can you please than explain what is the difference in acting on a ket $\left| \psi \right\rangle$ with an operator $\hat{x}$ like this $\hat{x}\left|\psi\right\rangle$ or calculating the inner product $\left \langle x | \psi(t) \right \rangle$ - if possible, could you provide a physicall interpretation (after 3rd you have misstyped an inner product i think. There is a $|$ missing i think). |
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1d |
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Some Dirac notation unclarities 1st: You never used symbol $\Psi$. Why is so? Is this common praxis? 2nd: Does a $\left|\psi\right\rangle$ in a $\hat{H}\left|\psi \right\rangle = W \left|\psi \right\rangle$ means a time independant wave function? 3rd: I allso thought that if i act with a position operator $\hat{x}$ on a ket $\left|\psi(t)\right\rangle$ i can denote this as $\hat{x}\left| \psi(t)\right\rangle$ why did you write this as $\left\langle x | \psi(t)\right\rangle$? 3rd Does $\left\langle x | \psi(t)\right \rangle$ denote a wavefunction represented in position space/basis? Explain please in an edit. |
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2d |
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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. |
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2d |
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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 :) |
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2d |
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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 $\left|B\right\rangle$? |
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2d |
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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! |
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2d |
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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. |
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May 12 |
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Some Dirac notation explanations Thank you i think i now understand a bit more. |
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May 11 |
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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 ... |
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May 10 |
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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: $\langle O\rangle = \int |\psi\rangle^\dagger O|\psi\rangle d x$ ? |
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May 10 |
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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. |
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May 9 |
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QM formalism is one big confusion - lack of geometrical explaination with images Thank you. It seems i complicate too much. |
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May 8 |
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QM formalism is one big confusion - lack of geometrical explaination with images And 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. |
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May 8 |
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QM formalism is one big confusion - lack of geometrical explaination with images So 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? |
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May 7 |
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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... |
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May 7 |
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Energy eigenvalues of a Q.H.Oscillator with $[\hat{H},\hat{a}] = -\hbar \omega \hat{a}$ and $[\hat{H},\hat{a}^\dagger] = \hbar \omega \hat{a}^\dagger$ Does this mean that eigenvalue in Hilbert space is equivalent of an expectation value??? |
