数学代写|PHYS451 Quantum mechanics

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PHYS451课程简介

Prerequisites, PHYS 250 and PHYS 201 or PHYS 207. Students learn the basic principles of quantum mechanics, which describe atomic, molecular, and nuclear physics. Students learn general concepts, methods, analytic tools and advanced problem-solving skills. Topics include evidence for and origins of quantum mechanics, mathematical background, the postulates of quantum mechanics, one-dimensional systems, quantization of angular momentum, and three-dimensional quantum systems including the hydrogen atom. Letter grade with Pass/No Pass option. (Offered fall semester.) 3 credit

Prerequisites 

Course Description
In this course, students learn the basics of non-relativistic quantum mechanics. The course introduces the concept of the wave function, its interpretation, and covers the topics of potential wells, potential barriers, quantum harmonic oscillator, and the hydrogen atom. Next, a more formal approach to quantum mechanics is taken by introducing the postulates of quantum mechanics, quantum operators, Hilbert spaces, Heisenberg uncertainty principle, and time evolution. The course ends with topics covering the addition of angular momenta, spin, and some basic aspects of many-body quantum mechanics. The course will include two lectures per week accompanied by a recitation.

PHYS451 Quantum mechanics HELP（EXAM HELP， ONLINE TUTOR）

The total probability should be equal to unity, i.e., the sum of all possible probabilities should be equal to one:
$$
\sum_{k=0}^{\infty} P(k)=1
$$
Substituting the given probability distribution function into the above equation, we get:
$$
\sum_{k=0}^{\infty} N \frac{\mu^k}{k !}=1
$$

Using the Taylor series expansion of $\$ \mathrm{e}^{\wedge}{\backslash m u} \$$, we can write:
$$
e^\mu=\sum_{k=0}^{\infty} \frac{\mu^k}{k !}
$$
Multiplying both sides by $\$ N \$$ and substituting $\$ \backslash$ sum__k=0 $}^{\wedge}{\backslash$ infty $}$ $\backslash \operatorname{frac}\left{\backslash m u^{\wedge} k\right}{k !} \$$ with $\$ e^{\wedge}{\backslash m u} \$$ we get:
$$
N e^\mu=1
$$
Therefore, the normalization factor $\$ N \$$ is:
$$
N=\frac{1}{e^\mu}=e^{-\mu}
$$

The average value of $\$ k \$$ is given by the first moment of the probability \& $Q$ distribution function:
$$
\langle k\rangle=\sum_{k=0}^{\infty} k P(k)=\sum_{k=0}^{\infty} k N \frac{\mu^k}{k !}=N \mu \sum_{k=1}^{\infty} \frac{\mu^{k-1}}{(k-1) !}=N \mu \sum_{k=0}^{\infty} \frac{\mu^k}{k !}=N \mu e^\mu
$$
Substituting the value of $\$ N \$$ from the previous question, we get:
$$
\langle k\rangle=\mu
$$

The average value of $\$ k^{\wedge} 2 \$$ is given by the second moment of the probability distribution function:
$$
\left\langle k^2\right\rangle=\sum_{k=0}^{\infty} k^2 P(k)=\sum_{k=0}^{\infty} k^2 N \frac{\mu^k}{k !}=N \mu^2 \sum_{k=2}^{\infty} \frac{\mu^{k-2}}{(k-2) !}=N \mu^2 \sum_{k=0}^{\infty} \frac{\mu^k}{k !}=N \mu^2 e^\mu
$$
Substituting the value of $\$ N \$$ from the previous question, we get:
$$
\left\langle k^2\right\rangle=\mu^2+\mu
$$

The variance of $\$ k \$$ is given by the difference between the second moment and the square of the first moment:
$$
\operatorname{Var}(k)=\left\langle k^2\right\rangle-\langle k\rangle^2=N \mu^2 e^\mu-(\mu)^2=\mu^2+\mu-\mu^2=\mu
$$
The standard deviation of $\$ \mathrm{k} \$$ is the square root of the variance:
$$
\sigma_k=\sqrt{\operatorname{Var}(k)}=\sqrt{\mu}
$$

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