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Roman P.
Physicist/Mathematician for 10 years with few years of tutoring experience
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Linear Algebra
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Question:

Using the Spectral Theorem for linear operators, find if the operator $(A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$) is diagonalizable.

Roman P.

According to the Spectral Theorem, the complex-valued linear operator is diagonalizable if it is normal (i.e. commutes with its adjoint). The adjoint is $(A^\dagger = \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}$) hence $( [A^\dagger, A] = A^\dagger A - A A^\dagger= \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}$) From the calculation, one sees that the operator is not normal, hence it is non-diagonalizable.

Calculus
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Question:

Find the volume of a ball of radius $$R$$ using volume integral in spherical coordinates.

Roman P.

Let us use spherical coordinates $$(r,\theta,\phi)$$. Cartesian coordinate of the radius vector in terms of spherical coordinates is $$(r \sin \theta \cos \phi, r\sin \theta \sin \phi , r \cos \theta)$$. The volume element in spherical coordinates is $$J = \left\| \frac{\partial (x,y,z))}{\partial(r,\theta,\phi)} \right\| = r^2 \sin\theta$$. The ranges for $$r,\theta,\phi$$ which cover the whole ball are $$0\leq r \leq R$$, $$0\leq \theta \leq \pi$$, $$0 \leq \phi \leq 2 \pi$$. Therefore, the volume is $$V = \int_{0}^{2\pi} d \phi \int_{0}^{\pi} \sin\theta d\theta \int_{0}^{r} r^2 dr$$. Because of a particular form of the volume element, it factors out each integral with respect to different variables. Evaluating simple table integrals in the last expression, one obtains well known formula $$V = \frac{4}{3} \pi R^3$$.

Physics
TutorMe
Question:

Is $$|\psi> = \frac{1}{\sqrt{3}} |0> + \sqrt{\frac{2}{3}}|1>$$ a valid qubit state? In case it is, what is the probability of the outcome 0 after the measurement in z basis?

Roman P.

In order for a qubit state to be valid, it has to be normalized to unity. Hence, one has to check if $$<\psi|\psi>=1$$ holds. Since $$|0>, |1>$$ form an ONB (orthonormal basis), the cross terms in the inner product vanish and $$<\psi|\psi>=\frac{1}{3} + \frac{2}{3} = 1$$. This is a valid qubit state. The probability of the outcome $$|\phi>$$ when the qubit is in state $$|\psi>$$ is $$p(\phi)=|<\psi|\phi>|^2$$, or in terms of the projection operator $$P_{|\phi>}$$, $$p(\phi)=<\psi | P_{|\phi>} | \psi>$$. Therefore, $$p(0) = |<0 | \psi>|^2 = \frac{1}{3}$$.

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