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3

As stated, $\mathbf{n}$ is a unit vector and $n_x$, $n_y$ and $n_z$ are its cartesian components. $\mathbf{n}$ is just a vector pointing in an arbitrarily direction with magnitude 1. Taking $\mathbf{n} \cdot \mathbf{\sigma}$, we have \begin{equation} \mathbf{n} \cdot \mathbf{\sigma} = n_x\sigma_x + n_y \sigma_y + n_z \sigma_z \\ = n_x \left(\begin{array}{cc} ...


3

What you really want to know are the definitions of the $\sigma_i$ --- these are the Pauli matrices: $$ \sigma_x = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \qquad \sigma_y = \begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} \qquad \sigma_z = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}$$ Hopefully you can see now how the equation ...


5

This is the same notation that you'll find in Weinberg's books. $$(\psi, \chi)$$ is the inner product of the two states $\psi$ and $\chi$, and corresponds to $$\langle \psi \mid \chi \rangle$$. So, the above corresponds literally to $$ \frac{1}{\sqrt{\langle \psi_k \mid \psi_k \rangle}} \left| \psi_k \right>$$ This new object is just the normalized ...


5

Canonically, the wedge product is distinct from the cross product and should not be confused, however in three dimensions they are inextricably linked. The wedge product (or outer product) comes from exterior algebra, first due to Grassmann who generalized vector products to arbitrary dimensions. This would later be extended by Clifford into the Clifford ...


7

The wedge product has its roots in exterior algebra. Exterior algebra lets you talk about objects like planes or volumes as algebraic elements of their own, separate from ordinary vectors, but still obeying the same notions of being "vectors" in their own vector spaces. The wedge product of two vectors is a bivector, and many concepts you may have been ...


2

Close, but not quite. Since quantum mechanics deals in probabilities, it is necessary to "normalize" the state in order to use it in later calculations. The most general state for the two-state quantum system you're considering would be \begin{equation} |\psi\rangle = \alpha\,|0\rangle+\beta\,|1\rangle \end{equation} where the quantities $\alpha$ and ...


2

Yes, that is correct. A more general form of the superposition of the stationary state would be $$a|0\rangle + b|1\rangle$$ where $a,b$ describes the probability of each state. The state : $$|0\rangle + |1\rangle$$ assumes that the state $|0\rangle$ and $|1\rangle$ are equally probable.


0

Yes, the reason for this is that in maths $\nabla$ is often used as the vector differential opertator. This is a vector. When typesetting the convention to denote a vector is bold text e.g. $\bf{x}$. However for handwriting you can't really write bold font so other conventions are needed. Common ones are putting an arrow over as in your example or ...


3

Yes, there are sometime different conventions for indicating vectors in hand-writing and printing. Yes, overset arrows in handwriting and boldface in printing is one of those conventions. No, it is not the only convention. Yes, you should familiarize yourself with the most common conventions in your sub-discipline. Yes, you should read the section on ...


0

If $L(t)$ were a scalar valued function of time then, by the product rule, we have $$\frac{d}{dt}L^2(t) = \frac{d}{dt}\left(L(t) \cdot L(t) \right) = \frac{dL}{dt} \cdot L(t)+ L(t) \cdot\frac{dL}{dt} = 2L(t)\cdot\frac{dL}{dt}$$ Due to linearity, this holds for ordinary vector valued functions of time since $$L^2 = \vec L \cdot \vec L$$ Thus ...


0

This is true for any vector quantity from a finite-dimensioned vector space that uses the standard definition of the inner product. Let $\mathbf u$ and $\mathbf v$ be elements of the vector space $\mathbb R^N$ with inner product $\mathbf u \cdot \mathbf v = \sum_{i=1}^N u_i v_i$. Then $\mathbf u \cdot \mathbf u = \sum_{i=1}^N {u_i}^2$. Using the standard ...


1

It is the easiest to think of this problem in the component form. Let's say the vector $\vec L$ is in a 3 dimensional space (that is usually what we use). So in component form, i.e., written as a 3 by 1 matrix, $\vec L=(L_1,L_2,L_3)$. What is the left-hand-side of your expression? It is $\vec L\cdot\frac{\mathrm d\vec L}{\mathrm dt}=L_1\cdot\frac{\mathrm ...


1

From math and the power rule: $\dfrac{d(x^2)}{dx} = 2x$ And we assume that L is a function of time: $\vec{L} = \vec{L(t)}$. To refresh you on the chain rule: if x were a function of time, then $\dfrac{d(f(x))}{dt} = \dfrac{d(f(x))}{dx} * \dfrac{dx}{dt}$. Back to math: $\dfrac{d(x^2)}{dx}$ is actually $2x\dfrac{dx}{dx}$ if you apply said chain rule. ...


5

This is notation for the imaginary part of a complex number. It is a fraktur letter I, and its counterpart for the real part is a fraktur letter R. Thus, if $z=x+iy$ and $x,y$ are real, one writes $$ \mathfrak{R}\,z=x\ \ \text{ and }\ \ \mathfrak{I}\,z=y. $$ A good chart of the fraktur alphabet is in this Yale resource, which includes handwriting guidance, ...


2

If it is actually the imaginary part of a complex variable, then just write $Im[\cdot]$ instrade of the curly character.


1

For me in my own field (optics, where one most often comes across it in engineering considerations), "amps" is common spoken usage, particularly for compound words such as milliamp or microamp. For written usage, I'm afraid I like to see the full name - it is, after all, recalling a very great man of science André-Marie Ampère. Even so, curiously, the SI ...


0

Well, amp is also used as a short form of "amplifier". I highly suggest to use the full name of the unit or its symbol, i.e. Ampere or A. :)


11

Technically, apparently, your teacher is correct. BIPM and NIST In the official brochure from the Bureau international des poids et mesures (BIPM, the keepers of SI units) in §5.1 Unit symbols we find: It is not permissible to use abbreviations for unit symbols or unit names, such as sec (for either s or second), sq. mm (for either mm2 or ...


13

If I saw the word "amp" written as such in a paper in my field (astrophysics) it would strike me as a bit informal. I would expect to see the full "ampere" written. That said, it is rare to actually write out the full name of a unit; usually it follows a number and is given its standard abbreviation. When abbreviated to e.g. "$5\ \mathrm{A}$", I would ...


1

According to the Wikipedia page, amp is acceptable, but is not a correct SI unit. I think your instructor is being thorough and making certain that you know the correct term to use.


1

All the possible states for your systems are encoded as rays in a Hilbert space. For expample, imagine that we are interested in the spin direction of an electron. The Hilbert space for this system is $\mathcal{H}=\mathbb{C}^2$. We can take an orthonormal basis, for example, the $z$ component of the spin, which it would be represented by the orthonormal ...


1

The unitary matrix as given represents an operation on the Hilbert space of states. It could be a measurement, it could be a symmetry, it could be time evolution, or something else. It is called "randomizing" here, because it transforms a state of definite spin (the $\lvert 0 \rangle = \left(\begin{matrix} 1 \\ 0 \end{matrix}\right) $ state) into a state of ...


1

Often times, an operation is something that changes the state of the system, like a measurement. But it can be other things that will change the state of the system.


6

What Goldstein means by $\nabla_iV_i$ is $$ \nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i $$ which is indeed a vector. Here, $\mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i})$ is the position of the $i$th particle (with respect to the origin), ...


5

This is typical to see in situations where the potential is a function of the coordinates of more than one particle: $$ V=V(\mathbf r_1,\ldots,\mathbf r_N)=V(x_1,y_1,z_1,\ldots,x_N,y_N,z_N). $$ The force produced by such a potential on the $i$th particle is the gradient of this function with respect to that particle's coordinates, while holding all the other ...


3

The $E$ he is refering to is the Young's Modulus, which is the elastic modulus that tells you how the tensile stress for a wire relates to its extensional strain, i.e. $$ \text{stress} = E \text{ strain} $$ in a thin wire. In the experiment he is summarizing in the graph, you pull a thin wire until it breaks, while measuring the stress on each end. The ...


0

$T, W, P$ $\to$ $T, O, I$ $C_n \to C_n$ $D'_n \to D_n$


1

The algebraic rule that if $xy=0$ then $x=0$ or $y=0$ does hold for tensor products: if you had $a^x b_y = 0$ you could conclude that $a^x = 0$ or $b_y = 0$. But there's no such rule for contractions like $a^x b_x$. For example suppose $a$ and $b$ written out in components were $(1,0)$ and $(0,1)$ respectively. You can't just move a variable to the other ...



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