I need (as a part of one exercise) to find commutator between $\hat{x}^2$ and $\hat{p}^2$ and my derivation goes as follows:  $$[\hat{x}^2,\hat{p}^2]\psi = [\hat{x}^2\hat{p}^2 - \hat{p}^2\hat{x}^2]\psi = - \hbar^2 x^2 \cdot \psi'' + \hbar^2 \frac{\partial^2}{\partial x^2}(x^2 \cdot \psi)$$  Now:  $$\frac{\partial}{\partial x}(x^2 \cdot \psi) = 2x\cdot \psi + x^2 \cdot \psi'$$  $$\frac{\partial}{\partial x}(2x\cdot \psi + x^2 \cdot \psi') = 2 \cdot \psi + 2x \cdot \psi' + 2x \cdot \psi' + x^2 \cdot \psi''$$  And then  $$[\hat{x}^2,\hat{p}^2]\psi = (2 \hbar^2 + 4 \hbar^2 x \cdot \frac{\partial}{\partial x})\psi$$  So I can derive, that  $$[\hat{x}^2,\hat{p}^2] = 2 \hbar^2 \cdot (1 + 2\hat{x}\hat{p})$$  I can not found this derivation anywhere and wonder: am I correct? Can there be other way to derive this?  I can not deduce any physical meaning from it, so any subtle mathematical error may go unnoticed.
20618
I have been told that $$[\hat x^2,\hat p^2]=2i\hbar (\hat x\hat p+\hat p\hat x)$$ illustrates operator ordering ambiguity.  What does that mean? I tried googling but to no avail.
83729
let's consider $f: \mathbb R^2 \to \mathbb R, \quad (x,y) \mapsto xy$  We calcualte the partial-derivatives:  $\partial_x f=y$   $\partial_y f=y$  Now let's sitch to polar-coordinates, we get:  $f: (0,\infty)\times\mathbb R \to \mathbb R, \quad (r,\varphi) \mapsto r^2\sin \varphi \cos \varphi$  We calcualte the partial-derivatives:  $\partial_r f = 2r\sin\varphi\cos\varphi$   $\partial_\varphi f = r^2(-\cos^2\varphi+\sin^2\varphi)$  Now, If I calcualted the derivation in polra-coordinates, can I somehwo go back to eucledean coordinates? Just using $x=r\cos\varphi, y=r\sin\varphi$ won't do it, since if we take the partial-derivatives in polar coordinates we would get:  $2r\sin\varphi\cos\varphi=2y\cos\varphi$  $r^2\sin\varphi - r^2\cos^2\varphi = x^2-y^2$  I first expected to get $x$ and $y$ but that's bullshit since we aren't deriving in the same directions. So how could I now switch properly to eucleadean coordinates?
58532
We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.  As background there are plenty of mathematical schemes that can be used to define these operators. One of these is   where $H_0(a,b) = b+1$   (successor op) $H_1(a,b) = a+b$   (addition op) $H_2(a,b) = ab $   (multiplication op) $H_3(a,b) = a^b$   (exponentiation op) $H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times ) etc.    Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$  Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?  Thanks.
117710
I don't understand how to calculate this generalized two-point function or propagator, used in some advanced topics in quantum field theory, a normal ordered product (denoted between $::$) is subtracted from the usual time ordered product (denoted $T$):  $$\langle X^{\mu}(\sigma,\tau)X^{\nu}(\sigma',\tau')\rangle  ~=~     T ( X^{\mu}(\sigma,\tau)X^{\nu}(\sigma',\tau'))     ~-~ : X^{\mu}(\sigma,\tau)X^{\nu}(\sigma',\tau'):$$   My question is can the rhs of this propagator be derived or the meaning of the subtraction of the time ordered product explained and motivated in simple words?
97037
A form of the  is  $$     \left[-\frac{\hbar^2}{2m} \nabla^2 + V(\vec{r}, t)\right]\Psi = i\hbar \frac{\partial}{\partial t} \Psi $$  The bracketed term is of course the Hamiltonian, and the whole equation can hence be written as  $$     \hat{H}\Psi = i\hbar \frac{\partial}{\partial t} \Psi $$  My question is, is there a existing/named operator represented by $i\hbar \frac{\partial}{\partial t}$? That is, can we further rewrite the equation as  $$     \hat{H}\Psi = \hat{\Theta}\Psi $$  For some predefined operator $\hat{\Theta} = i\hbar \frac{\partial}{\partial t}$?
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?
78877
\documentclass[11pt]{article}  \begin{document}   The above equations on application of Laplace Transform become:  $$sL_1\hat{i}_{L1}(s) = D\cdot \hat{v}_{C1}(s) + D\cdot \hat{v}_{C2}(s) - D\prime \cdot \hat{v}_g(s) + (C_{C1} + V_{C2} - V_g + V_D)\cdot \hat{d}(s)$$ $$sL_2\hat{i}_{L2}(s) = -D\prime \cdot \hat{v}_{C1}(s) + D\cdot \hat{v}+{C2}(s) + D\prime \cdot \hat{v}_g(s) + (C_{C1} + V_{C2} - V_g + V_D)\cdot \hat{d}(s)$$ $$sC_1\hat{v}_{C1}(s) = -D\cdot \hat{i}_{L1}(s)+ D\prime \cdot \hat{i}_{L2}(s) + (-I_{L1} -I_{L2} - I_1)\cdot \hat{d}(s) -D\prime \cdot \hat{i}_l(s)$$ $$sC_1\hat{v}_{C1}(s) = D\prime \cdot \hat{i}_{L1}(s)- D\cdot \hat{i}_{L2}(s) + (-I_{L1} -I_{L2} - I_1)\cdot \hat{d}(s) -D\prime \cdot \hat{i}_l(s)$$ $$sL_l\hat{i}_l(s) = D\prime \cdot \hat{c}_{C1}(s) + D\prime \cot \hat{v}_{C2}(s) - D\prime \cdot \hat{v}_g(s) +(V_g - V_D - V_{C1} - V_{C2})\cdot \hat{d}(s) - R_l \cdot \hat{i}_l(s)$$  \end{document}   Error:   181 Underfull \hbox (badness 10000) in paragraph at lines 181--181 181 Overfull \hbox (22.63293pt too wide) detected at line 181
226
On page 19 of the textbook  (by James, Witten, Hastie and Tibshirani--it is freely downloadable on the web, and very good), the following is stated:  Consider a given estimate $$\hat{Y} = \hat{f}(x)$$ Assume for a moment that both $$\hat{f}, X$$ are fixed. Then, it is easy to show that: $$\mathrm{E}(Y - \hat{Y})^2 = \mathrm{E}[f(X) + \epsilon - \hat{f}(X)]^2$$ $$ = [f(X) - \hat{f}(X)]^2 + \mathrm{Var}(\epsilon)$$  It is further explained that the first term represents the reducible error, and the second term represents the irreducible error. I am not fully understanding how the authors arrive at this answer. I worked through the calculations as follows: $$\mathrm{E}(Y - \hat{Y})^2 = \mathrm{E}[f(X) + \epsilon - \hat{f}(X)]^2$$ This simplifies to $[f(X) - \hat{f}(X) + \mathrm{E}[\epsilon]]^2 = [f(X) - \hat{f}(X)]^2$ assuming that $\mathrm{E}[\epsilon] = 0$. Where is the $\mathrm{Var}(x)$ indicated in the text coming from? Any suggestions would be greatly appreciated.
101136
I need to prove that $(G_{1}\times G_{2})^{\prime} = G_{1}^{\prime} \times G_{2}^{\prime}$, where $G^{\prime}$ denotes the commutator subgroup of $G$ - i.e., $G^{\prime}=[G,G]$, the subgroup generated by all commutators of elements of $G$.   Recall that the commutator of two elements $x$ and $y$ in a group $G$, denoted $[x,y]$ is defined to equal $x^{-1}y^{-1}xy$.  I have posted my proof as an answer below. Could somebody please take a look at it and let me know if it's okay? If not, please let me know what I need to do in order to fix it. Thank you. :)    Edit: I have been informed that my answer given below actually shows only equality of the set of commutators, and that $G^{\prime}$ consists of products of commutators. So, this question is no longer a proof check, but I am asking specifically how to fix what I have written below in order to make it actually answer the thing I set out to prove. I am having a bit of trouble understanding some of the hints given me thus far, and am going to need more detail in any answers given in order for them to be helpful.
34579
I'm reading Sakurai's Modern QM right now and in the first chapter he states a number of conditions required for a translation operator: unitarity, $$T(\mathrm{d}\mathbf{x})T(\mathrm{d}\mathbf{x^\prime}) = T(\mathrm{d}\mathbf{x}+\mathrm{d}\mathbf{x^\prime}),$$ $$T(-\mathrm{d}\mathbf{x}) = T^{-1}(\mathrm{d}\mathbf{x}),$$ and going to 1 for small translations.   He then makes the ansatz (1.6.20 in the new version): $$ T(\mathrm{d}\mathbf{x}) = 1-i \mathbf{K} \cdot \mathrm{d}\mathbf{x} $$ and demonstrates that this form of $T$ meets the requirements and is generally the correct choice. Here, $\mathbf K =(\mathrm K_x,\mathrm K_y,\mathrm K_z)$ with each component hermitian.  I was hoping somebody could provide some motivation for this particular choice. Is there any reason why it is to be expected or natural? How would you come up with this form of the operator by yourself?
1291
Let $\phi$ be a scalar field and then I see the following expression (1) for the square of the normal ordered version of $\phi^2(x)$.   \begin{align} T(:\phi^2(x)::\phi^2(0):) &amp;= 2 \langle 0|T(\phi(x)\phi(0))|0 \rangle^2 \\ &amp;+ 4\langle 0|T(\phi(x)\phi(0))|0 \rangle:\phi(x)\phi(0): \\ &amp;+ :\phi^2(x)\phi^2(0): \tag{1} \end{align}  It would be great if someone can help derive the above expression - may be from scratch - and without outsourcing to Wick's theorem - and may be help connect as to why the above is related (equal?) to the Wick's theorem?    Isn't the above also known as OPE (Operator Product Expansion)? If yes, then is there at all any difference between OPE and Wick's theorem? Is there a systematic way to derive such OPEs?  Can one help extend this to Fermions?
16471
I am struggling to show that $$\frac{1}{\sqrt{2\pi}} \cdot \left(\int_{- \infty}^{\infty} e^{itx} \cdot e^{-\frac{1}{2}x^2 } dx \right) = e^{-\frac{1}{2}t^2}$$  In the same paper we defined  $$F(t):=\left(\int_0^t e^{-x^2}\right)^2 \quad \text{ and} \quad  G(t):=\left( \int _0^1\frac{e^{-t^2(1+x^2)}}{1+x^2} \right)$$  I showed that both functions are differentiable with  $$F'= 2e^{-t^2} \cdot \int_0^t e^{-x^2}$$ $$G' = -2te^{-t^2} \cdot \int_0^1 e^{-x^2t^2}$$  It follows that $F' + G' = 0$ and $F + G = \frac{\pi}{4}$  With that I showed that $\int_{-\infty}^{\infty} e^{-x^2} = \sqrt{\pi}$.  Any help is appreciated.
26750
I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of how to proceed.  If A and B commute then $[A,B] = ABA^{-1}B^{-1} = e$ where e is the identity element of the group.  $$\therefore AB=BA$$  $$n=1; [A,B^1] = (1)B^0[A,B] = e$$ This statement is certainly true. however moving on to $n = 2$ I find...  $$[A,B^2] = AB^2A^{-1}B^{-2} = ABBA^{-1}B^{-1}B^{-1} = BBAA^{-1}B^{-1}B^{-1}$$  Where in the last step I have used the fact that A and B commute to rearange the terms. However, it is plain to see that this last term simply reduces to the identity as well and for the n = 2 case we have:  $$[A,B^2] = e \ne 2B[A,B] = 2Be = 2B$$  Clearly I have assumed something I shouldn't have. The fact that there is a multiplicative factor of n implies I should be adding things, but I thought if I kept it as general as possible, the answer should just fall out naturally. I don't want an answer please, only guidance.
4998
I'm asked to show that  $$\frac{d(\hat{A}\hat{B})}{d\lambda} ~=~ \frac{d\hat{A}}{d\lambda}\hat{B} + \hat{A}\frac{d\hat{b}}{d\lambda}$$ With $\lambda$ a continuous parameter. Should I use the definition $$\frac{d\hat{A}}{d\lambda} ~=~ \lim_{\epsilon \to 0} \frac{\hat{A}(\lambda + \epsilon) - \hat{A}(\lambda)}{\epsilon}$$ applied to $\hat{A}\hat{B}$ like $$\frac{d(\hat{A}\hat{B})}{d\lambda} ~=~ \lim_{\epsilon \to 0} \frac{\hat{A}(\lambda + \epsilon)\hat{B}(\lambda + \epsilon) - \hat{A}(\lambda)\hat{B}(\lambda)}{\epsilon}$$ and do some algebra to get the RHS of the first equation, or I'm missing something? Another interesting derivative to pay attention to is: $$\frac{d}{d\lambda}\exp(\hat{A}(\lambda) )~?$$
104966
For the definition of the momentum operator $$\hat{P } = -i \hbar \nabla$$ in quantum mechanics, as I understand you can derive this by either considering a more general definition of momentum, i.e. 'canonical momentum' which is an operator and then apply this operator to wave functions. This is shown here in this . We can alternatlively start with translations and use the face that the momentum operator is a generator of translations as is done here in this .   What I am interested in is what is the more fundamental derivation for the position operator: $$\hat{X} = x.$$ To this point I have considered that the motivation for defining position operator is from the definition of the expectation value $$\langle x \rangle = \int dx x |\psi(x)|^2 = \langle \psi|x | \psi \rangle$$ where $| \psi \rangle$ is normalised. Is this the full extent of the motivation or the main point to consider? Is there another motivation for the position operator or is it taken from this motivation and confirmed by experiments to be acceptable?
38398
I am a bit confused by the concepts of . In all the courses I am doing at the moment we do transformations of the form: $$ \phi(x) \rightarrow\phi'(x') = \phi(x) $$ and $$ \partial_{\mu}\phi(x)\rightarrow\partial' _{\nu}\phi(x') = \frac{\partial x^{\alpha}}{\partial x' ^{\nu}}\partial_{\alpha}\phi(x) $$ This is all perfectly clear to me. However, I am reading Peskin and Schoder at the moment, and they adapt an "active" point of view (their words), such that the above transformations are: $$ \phi(x) \rightarrow\phi'(x) = \phi(\Lambda^{-1}x) $$ and $$ \partial_{\mu}\phi(x)\rightarrow\partial _{\mu}(\phi(\Lambda^{-1}x)) = (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x). $$ I don't understand how to interpret this and especially how to derive the second equation.
71857
Help me find $\hat{B^\dagger}$, when we know that $$\hat{B}=i\frac{d}{dr}$$ with the condition that $\hat{B}$ is defined in spherical coordinates. My approach: $$ \langle\psi|\hat{B}\psi\rangle=\int_{0}^{\infty} \psi^* i\frac{d}{dr} \psi dr=\psi^*\psi|_{0}^{\infty}-\int_{0}^{\infty}\psi i\frac{d}{dr}\psi^*dr=\langle\hat{B}\psi|\psi^*\rangle$$ And so I get $ \hat{B^\dagger} = -i\frac{d}{dr} $. Could someone confirm if this is correct?