Advanced Quantum Mechanics Lecture 1

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Recommends Zainy · 2015-10 · Physics Stack Exchange · on “Trying to understand inner product notation”

I I'm taking a QM course and I'm trying to make sense of why observables are sometimes conjugated for no apparent reason in their inner products. Right now I'm watching Dr. Susskind's lecture on youtube ( [this video] ) and around 13:16 he takes the inner product of the ket wavefunction $$ \mid\psi(x)\rangle $$ and the state representing the particle at $x$ , $$\mid x \rangle $$ to make $$\langle x \mid \psi \rangle $$ Doesn't this conjugate the $x$ term? My question is why all of a sudden does he conjugate the position vector to complete the inner product? Right before that he took the inner product with two different state vectors, conjugating the $ x' $ term but not the $ x $ term. I guess I'm

Recommends Noumeno · 2020-11 · Physics Stack Exchange · on “How Rising and Lowering Operators directly imply quantization?”

Studying quantum angular momentum from my lectures and also from this [this video] useful collections of lectures (from Leonard Susskind) I have of course stumbled upon rising and lowering operators (a.k.a ladder operators). I am completely fine with the definition of those operators and with the fact that applying them allows us to discover new eigenstates of the angular momentum, with different eigenvalues associated to them. But from this my lectures go on to state that the existence of this operators alone allows us to prove that: Angular momentum is quantized. We can find all eigenstates of angular momentum by applying the ladder operators to a known eigenstate of it. In both my lectures and

Recommends MichaelW · 2023-11 · Physics Stack Exchange · on “What is the deeper relation between symmetry and the Hamiltonian?”

Note: This question [LINK] is the same as mine. After having studied physics decades ago I recently reviewed some basics of symmetry in quantum physics. As a more experimental physicist I never understood deeply, what symmetries have to do with the Hamiltonian operator $H$ . For instance it is always stated, that $H$ is the generator of time evolution, so (I put it down simplified and also set $\hbar=1$ ) $$| \psi(t+dt)\rangle = (1- iHdt)\, | \psi(t)\rangle $$ Similarly, $$\psi(x+dx)= \psi(x) + dx \cdot \frac{\partial }{\partial x} \psi(t) = (1- i p \cdot dx) \, \psi(t)$$ with $p$ as the momentum operator. Why is there a symmetry , when $$[p, H]=0.$$ What has time evolution to do with a shif

Cites Timaeus · 2015-10 · Physics Stack Exchange · on “Trying to understand inner product notation”

I'm taking a QM course and I'm trying to make sense of why observables are sometimes conjugated for no apparent reason in their inner products. There is a lot of confusion even already. Let's start with the basics. There are vectors and operators and an inner product. If everything was finite dimensional the vectors would be complex column vectors, the operators would be complex square matrices and the inner product would be taken by transposing one vector and then conjugating every term in it and then doing regular matrix multiplication. Note right away that unlike regular 3d space this new inner product is different. Firstly it is complex, secondly you no longer get that it is symmetric. T