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Here is what I understand: if you have a particle at state $|x \rangle$, active translating it by $a$ means moving the particle to state $| x + a \rangle$. Passive transformation means you keep the particle in the same place, and change the coordinate by new variable $x = x' + a$ (note that the coordinate system is translated backwards $-a$). I am not very ...

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The difference between active and passive transformations is that in active transformations you are transforming the state, where as in passive transformations you are transforming the operator. If you have a transformation on a state vector by a unitary operator of the form $$|\, \psi'\rangle \to U| \,\psi\rangle$$ From this we can show that the expectation ...

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Both $$\sum_i |i\rangle \langle i |$$ and $$\sum_j |j\rangle \langle j |$$ are summations over basis vectors. The indices $i,j$ run over the same values – values of indices that identify the basis vectors in the same basis (set of vectors) – but the particular values of the indices $i,j$ are independent. Can you calculate how much is the expression below?...

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It is better to see an explicit example. Consider a vector $\vec v\in\mathbb R^3$. There are three linearly independent subspaces of $\mathbb R^3$ generated by the unit vectors $\vec e_i$, $i=1,2,3$. If you act with the most general block diagonal $3\times 3$ matrix on $\vec v=\sum v_i\vec e_i$ you get $$\begin{bmatrix} a&0&0\\ 0&b&c\\ 0&... 0 One example for a quantum mechanical system is "one free electron in 3-dimensional space". The state "the electron is at the point \vec x_1" is described by one wave function \psi_1(\vec x), the state "the electron is at the point \vec x_2" is described by another wave function \psi_2(\vec x), and a third wave function \psi_3(\vec x) \sim \psi_1(\... 0 The variables in your experiment were : the mass of water in the calorimeter m_w (independent variable) and the temperature rise in the calorimeter \Delta T = T_{final} - T_{initial} (dependent variable). Your equation can be written as : (m_w c_w + m_c c_c)\Delta T = m_s c_s (\Delta T' - \Delta T) \frac{m_w}{m_s} + \frac{m_c}{m_w} \frac{c_c}{c_w} = \... 6 I'm not sure if it helps you with your students, but maybe gives you some background: I guess the underlying reason for orthogonal basis vectors is that you are implicitly using a euclidean metric that will just have diagonal values. These would e.g. be$$g_\mathrm{\mu\nu, ~euclidean}=I=\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} \...

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We don't always use orthogonal coordinate frames. For example working with three phase motors it's sometimes convenient to work with a three axis coordinate system in a plane. Convenience, simplicity set aside, the main reason we most often work with orthogonal reference frames is the concept of dimension. We can express an n-dimensional linear system as a ...

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No, an arbitrary operator does not represent a change of basis. And even those that can be used to perform changes of basis should not always be interpreted as such. A "change of basis" in a Hilbert space is usually meant to be a change from one orthonormal basis to another. The operators that map orthonormal systems to orthonormal systems are precisely the ...

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I can see why you think physicists might have some insight into such a problem, because of the way it has been presented as an electrical circuit. Unfortunately we have little to add to what the Mathematicians and Computer Scientists can tell you. We certainly have no tricks which they do not already know about. I think this is a mathematical or ...

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