I came across this expression:  $$\langle x\ |\ \psi\rangle=\psi(x)$$  How can it be justified? I understand the LHS as an inner product, and the RHS just as a function of the parameter $x$.
120880
I have come across the equation which comes out of the nothing  in Zettili's book Quantum mechanics concepts and applications p. 167:  $$\psi(\vec{r},t) ~=~ \langle \vec{r} \,|\, \psi(t) \rangle.$$   How do I know that I get function of position and time if i take an inner product of a state vector with an vector $\vec{r}$? Where is the proof?
119242
I am trying to understand how the correlation function in John Bell's paper on EPR is derived for a spin singlet state $|{\Psi}\rangle$. This is defined to be   $$ \langle{\Psi}|\left(\bf{\sigma}\cdot\bf{a}\right)\left(\bf{\sigma}\cdot\bf{b}\right)|{\Psi}\rangle=-\bf a \cdot\bf b. $$ I tried to compute it explicitly by using the Pauli matrices, but was unable to derive the scalar product of the two direction vectors.    Edit  One attempt to prove this is by using  $$ \langle{\Psi}|\left(\bf{\sigma}\cdot\bf{a}\right)\left(\bf{\sigma}\cdot\bf{b}\right)|{\Psi}\rangle=\langle{\Psi}|(\bf {a} \cdot\bf{ b})\bf I +i(\bf a \times\bf b)\cdot\sigma|{\Psi}\rangle $$ However, I am not able to derive the minus sing or get rid of the cross product term either. See the comment section
36068
I have an infinite basis which associates with each point, $x$, on the $x$-axis, a basis vector $|x\rangle$ such that the matrix of $|x\rangle$ is full of zeroes and a one by the $x^{\mathrm{th}}$ element. The book on Quantum Mechanics by Shankar says that the inner product between a basis vector and itself is not one, why not? Why can't these basis vectors be normalized to one, only to the ?
11306
I have been going through some problems in Sakurai's Modern QM and at one point have to calculate $\langle \alpha|\hat{p}|\alpha\rangle$ where all we know about the state $|\alpha\rangle$ is that $$\langle x|\alpha\rangle=f(x)$$ for some known function $f$. ($|\alpha\rangle$ is a Gaussian wave packet.) Sakurai says that this is given by:  $$\langle p\rangle = \int\limits_{-\infty}^{+\infty}\langle\alpha|x\rangle\left(-i\hbar\frac{\partial}{\partial x}\right)\langle x|\alpha\rangle dx.$$  I am wondering how we get to this expression. I know that we can express   $$|\alpha\rangle =\int dx|x\rangle\langle x|\alpha\rangle$$  and  $$\langle\alpha|=\int dx\langle\alpha|x\rangle\langle x|,$$   so my thinking is that we have:  $$\langle\alpha|\hat{p}|\alpha\rangle =\iint dx dx'\langle\alpha|x\rangle\langle x|\hat{p}|x'\rangle \langle x'|\alpha\rangle, $$  and if we can 'commute' $|x\rangle$ and $\hat{p}$ this would become: $$\iint dxdx'\langle\alpha |x\rangle\hat{p}\langle x|x'\rangle \langle x'|\alpha\rangle,$$  which is the desired result as $$\langle x|x'\rangle=\delta(x-x').$$ Is this approach valid?  I think my question boils down to: Does the operator $\hat{p}$ act on the basis kets $|x\rangle$ or on their coefficients? In the latter case, if we had some state $|\psi\rangle = |x_0\rangle$ for some position $x_0$, then would we say that for this state $$\langle p\rangle =\langle x_0|\left(-i\hbar\frac{\partial}{\partial x}\right)|x_0\rangle = 0$$ as the single coefficient is $1$ and the derivative of $1$ is $0$?
27781
I have a question about deriving the coordinate representation of momentum operator from the commutation relation, $[x,p]= i$.  One derivation (ref W. Greiner's Quantum Mechanics: An Introduction, 4th edi, p442) is as following: $$ \langle x|[x,p]|y \rangle = \langle x|xp-px|y \rangle = (x-y) \langle x|p|y \rangle. $$ On the other hand,  $ \langle x|[x,p]|y \rangle = i \langle x|y \rangle = i \delta (x-y)$. Thus $$ (x-y) \langle x|p|y \rangle = i \delta(x-y). \tag{1} $$  We use $(x-y) \delta(x-y) = 0$. Take the derivative with respect to $x$; we have $\delta(x-y) + (x-y) \delta'(x-y) = 0$. Thus $$ (x-y) \delta'(x-y) = - \delta(x-y). \tag{2} $$  Comparing Eqs. (1) and (2), we identify $$ \langle x|p|y \rangle = -i \delta'(x-y). \tag{3} $$  In addition, we can add $\alpha \delta(x-y)$ on the right-hand side of Eq. (3), i.e. $$ \langle x|p|y \rangle = -i \delta'(x-y)  + \alpha \delta(x-y), $$ and $[x,p] = i$ is still satisfied. We can also add $$ \frac{\beta}{\sqrt{|x-y|}}\delta(x-y)$$ on the RHS of Eq. (3).  Here $\alpha$ and $\beta$ are two real numbers.  My question is, what is the most general expression of $\langle x|p|y \rangle$? Can we always absorb the additional term into a phase factor like Dirac's quantum mechanics book did?  Thank you very much in advance.
86622
What would be the proof for $\langle q| p \rangle = e^{ipq}$?  Is it derived from canonical commutation relation?  ($|q \rangle $ represents the position eigenstate, while $|p \rangle$ represents the momentum eigenstate.)
111159
It is stated that if $\psi(x_0)=\psi'(x_0)=0$ at some $x_0$, then $\psi(x)=0,\forall x$.  Hence, we cannot have the wavefunction and its derivative vanishing at the same point.  My question is, how to prove the above statement?
81369
I was recently reading Griffiths' Introduction to Quantum Mechanics, and I stuck upon a following sentence:     but $\Psi$ must go to zero as $x$ goes to $\pm\infty$ - otherwise the wave function would not be normalizable.   The author also added a footer: "A good mathematician can supply you with pathological counterexamples, but they do not arise in physics (...)".  Can anybody give such a counterexample?
113328
This relates a little bit to my previous question (), but it's different in a way and more historical.  One of the postulates of quantum mechanics is:     To every observable in classical mechanics there corresponds a linear, Hermitian operator in quantum mechanics ().   However, absent in the list of postulates is what these operators are.  So is the mathematical expression for each individual operator also a postulate that's not listed, or are they derivable from other axioms?  Specifically, where did  expressions come from?  Maybe the question can better be answered historically.  If I had to guess, I'd say it began with the wavefunction.  Some scientists thought perhaps the squared modulus of the wavefunction corresponded to charge distribution or something, but Born was the one who got it right.  He essentially made a guess that the squared modulus corresponded to the probability of finding a particle at a certain location.  What's funny is that he initially wrote down the wrong expression in his paper (), and his paper even got rejected by the first journal he submitted it to, but the right expression is in a corrective footnote.    But again, Born simply guessed what the squared modulus meant.  There was no procedure; he just thought of an idea and then experiment verified it to be correct.  Thus:  $$\langle\psi|\hat{X}|\psi\rangle = \langle\psi|x\rangle\langle x|\hat{X}|x\rangle\langle x|\psi\rangle = \int d\vec{x}\ \psi^*(\vec{x})\ \psi(\vec{x})$$  So basically in the position basis, the position operator is simply $\hat{X} = x$, which seems pretty intuitive.  I believe the momentum operator came about next; and it was "derived" by fiddling around with the de Broglie relation and just substituting classical expressions for momentum into the Schrodinger equation (can't remember where I read that).  The fact that plugging these classical ideas into a quantum equation just happens to match experiment for momentum measurements seems like a pure coincident to me (it's along the same lines as trying to make the Schrodinger equation relativistic; one way of shoehorning relativity into it gives you the Dirac equation and a different way gives you the Klein-Gordon equation).  I imagine the spin operator came about by simple observation.  An electron goes up or electron goes down in a magnetic field.  Not hard to guess what the expression for that operator might be.  So anyway, am I on the right track here?  Have all of the operators come about simply by guessing?  And if I'm wrong, how did they actually come about?  How would I come up with a new operator for a new type of measurement?
46537
I'm trying to derive $\left\langle \frac{1}{r} \right\rangle = \frac{1}{n^2 a_0}$ (where $a_0$ is the Bohr radius) for the $| nlm\rangle$ state of hydrogen.  I've separated the radial and angular parts of the hydrogen wavefunction and split up the integration to yield  \begin{align}\left\langle \frac{1}{r} \right\rangle &amp;= \int R^*_{nl}(r) {Y^{m}_{l}}^*(\theta, \phi) \frac{1}{r} R_{nl}(r) Y_{l}^{m}(\theta, \phi) r^2 \sin \theta ~\mathrm dr ~\mathrm d\theta~\mathrm d\phi\\ &amp;= \int {Y^{m}_{l}}^*(\theta, \phi) Y_{l}^{m}(\theta, \phi) \sin \theta ~\mathrm d\theta ~\mathrm d\phi \int \limits_{0}^{\infty} r R^*_{nl}(r) R_{nl}(r) ~\mathrm dr\\ &amp;= \int \limits_{0}^{\infty} r R^*_{nl}(r) R_{nl}(r) ~\mathrm dr\end{align}  But from here, I'm not sure how to continue. Since $R_{nl}(r)$ is defined as $$R_{nl}(r) = \left( \frac{2 Z}{n a_0} r \right)^l \sum \limits_{k = 0}^{n - l - 1} a_k \left( \frac{2 Z}{n a_0} r \right)^k e^{-Zr/n a_0}$$ where $$a_{k + 1} = \frac{k + l + 1 - n}{(k + 1)(k + 2l + 2)}a_k$$  How do I deal with the sum inside the integral (especially a sum that requires recursion to compute coefficients)?  Have looked this  so far.
54908
Is it logically correct to assert that the expectation of the momentum $$\langle \hat p \rangle=0$$ for any bound state because it is bound to some finite region? What is the physical interpretation of the fact that $$\langle \hat p \rangle=0$$ in an energy eigenstate $\psi_n(x,t)$  but $$\langle \hat p \rangle\neq0$$ in some superposition state $$\psi(x,t)=c_m\psi_m(x,t)+c_n\psi_n(x,t)~?$$ Here $\psi_n(x,t)$ the eigenstates of the Hamiltonian, for example, in the problem of particle in a box (say).
101363
I am starting to learn about quantum mechanics and was wondering about the solutions to a question about a particle in an infinite potential well of length $L$.  The solutions are: $$\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{\pi n x}{L} \right)$$  And I was wondering what is the meaning of $n$. Isn't there supposed to be only one wave function? In that case, $\Psi = \sum_{n=1} ^\infty a_n \psi_n$, but what are the $a_n$? Or maybe it is probabilistic, and we don't know which $\psi_n$ the particle is currently in?
33691
How to show that $\mathbb R[x]/\langle x^2+1\rangle\simeq\mathbb C$ and $\mathbb R[x]/\langle x^2+1\rangle$ contains a zero of $x^2+1$      Due to the division algorithm, $\mathbb R[x]/\langle x^2+1\rangle=\{ax+b+\langle x^2+1\rangle:a,b\in\mathbb R\}$  Indeed I'm having problem to interpret, $\mathbb R[x]/\langle x^2+1\rangle$ contains a zero of $x^2+1.$ Does it mean $\exists$ $ax+b+\langle x^2+1\rangle\in\mathbb R[x]/\langle x^2+1\rangle$ such that $$(ax+b+\langle x^2+1\rangle)^2+(1+\langle x^2+1\rangle)=\langle x^2+1\rangle?$$
92246
For a given inner product space $V$, I would like to prove the inequality $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = 2(\|\mathbf{x}\|^2 + \|\mathbf{y}\|^2).$$  We see that $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = \langle\mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x}-\mathbf{y},\mathbf{x}-\mathbf{y}\rangle,$$  which by linearity $$=\langle\mathbf{x},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x},\mathbf{x}-\mathbf{y}\rangle - \langle\mathbf{y},\mathbf{x}-\mathbf{y}\rangle.$$  I've been trying to figure out how to go about this proof using linearity in the second argument of an inner product, but my textbook does not say that linearity necessarily holds in the second component. If it does, how would I go about the rest of this proof? If not, should I be potentially using some aspect of conjugate symmetry to prove this statement?
43963
In the Schrödinger equation, at any given time $t$ we should jointly add another sub equation, like $$||\psi_t(x)|| = 1$$ where $\psi_t(x) = \Psi(x,t)$, and then try to solve the two equations simultaneously. Why not? I know it does not yield, but I am always baffled, who is doing the normalization of the wave function? Observers, the system, the measuring process, God?
5833
I simply have no idea how this result is found $$I=\hbar  \sqrt{\ell(\ell+1)}.$$ The result seems to just be dumped in textbooks rather than explained. I can get the result that $I_z=\hbar m_j$. Please explain as clearly as possible how this is derived.  Also, any pointers towards any books/videos that may help me learn quantum mechanics would be appreciated.
117278
If $\|\cdot \|$ is the norm induced by the inner product $\langle,\rangle$, how to prove the following interesting inequality? $$\langle x,y\rangle(\|x\|+\|y\|) \leq\|x+y\|\,\|x\|\,\|y\|$$  This is a exercise in my textbook which is available only in portuguese called "Topologia e Análise no Espaço $\mathbb{R}^n$". The above inequality is obvious when $\langle x,y\rangle \leq 0$, but I don't know how to proceed to prove the other case.