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Homework 2 PHGN530 Statistical Mechanics Note: Please collaborate with your classmates heavily. This is very important homework for your learning. Due date before class: 09/20/2016 1. Importance of Liouville’s theorem in science (150 points total): Consider a simple harmonic oscillator in 2D with spring constant, k, and mass, m. Plot the following: a) y vs x at 0 t , 1 t s and 10 t s . (50 points) b) x p and y p at 0 t , 1 t s and 10 t s . (50 points) c) x p vs x and y p vs y at 0 t , 1 t s and 10 t s . (100 points) 2. Importance of Liouville’s theorem in science (400 points total): Consider 10 simple harmonic oscillators in 2D with spring constant, k, and mass, m. Plot the following: a) y vs x at 0 t , 1 t s and 10 t s . (100 points) b) x p and y p at 0 t , 1 t s and 10 t s . (100 points) c) x p vs x and y p vs y at 0 t , 1 t s and 10 t s . (200 points) Learning goal: Liouville’s theorem is one of the most fundamental results in physics and can be applied to both equilibrium and non-equilibrium situation. 3. Ergodicity: Starting from Liouville’s theorem, you can formulate different ensemble statistical mechanics by assuming that ergodicity is valid at equilibrium. Please search literature and find a recent paper where researchers have shown breakdown of ergodicity and write a short paragraph. (300 points) Learning goal: Assumption of ergodicity makes calculation easier, but it may not have to be valid strictly. As you a graduate student, you should be able to search, rethink, and come up with your own science. 4. Arrow of time or second law of thermodynamics: For macroscopic systems with large number of particles, entropy must increase for irreversible processes and remain same for reversible processes. But for smaller systems, it does not have to be true. You are not getting any homework on this now, but we will get back to this topic in October. For now, problem 4 is just for your information and not a problem really.

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  • Homework 2

    PHGN530 Statistical Mechanics

    Note: Please collaborate with your classmates heavily. This is very important homework for your learning.

    Due date before class: 09/20/2016

    1. Importance of Liouville’s theorem in science (150 points total): Consider a simple harmonic oscillator in 2D with spring constant, k, and mass, m. Plot the following: a) y vs x at 0t , 1t s and 10t s . (50 points) b) xp and yp at 0t , 1t s and 10t s . (50 points) c) xp vs x and yp vs y at 0t , 1t s and 10t s . (100 points)

    2. Importance of Liouville’s theorem in science (400 points total): Consider 10 simple harmonic oscillators in 2D with spring constant, k, and mass, m. Plot the following: a) y vs x at 0t , 1t s and 10t s . (100 points) b) xp and yp at 0t , 1t s and 10t s . (100 points) c) xp vs x and yp vs y at 0t , 1t s and 10t s . (200 points)

    Learning goal: Liouville’s theorem is one of the most fundamental results in physics and can be applied to both equilibrium and non-equilibrium situation.

    3. Ergodicity: Starting from Liouville’s theorem, you can formulate different ensemble statistical mechanics by assuming that ergodicity is valid at equilibrium. Please search literature and find a recent paper where researchers have shown breakdown of ergodicity and write a short paragraph. (300 points)

    Learning goal: Assumption of ergodicity makes calculation easier, but it may not have to be valid strictly. As you a graduate student, you should be able to search, rethink, and come up with your own science.

    4. Arrow of time or second law of thermodynamics: For macroscopic systems with large number of particles, entropy must increase for irreversible processes and remain same for reversible processes. But for smaller systems, it does not have to be true. You are not getting any homework on this now, but we will get back to this topic in October. For now, problem 4 is just for your information and not a problem really.

  • Statistical Mechanics Homework #2Alex WilhelmCollaborator: Kirsten Blagg (helped me catch up and explained the configuration of the first two problems after I returned from out of town)9/22/2016

    Problem #1Set up the system assuming identical springs:

    1) H = p2

    2m + 12 k q2 Here, p = px x + py y and q = x x + y y2) Hamilton’s equations of motion are: dpidt = -dHdqi & dqidt = dHdpi . These describe the motion of amechanical object in the Hamiltonian formalism.

    3) Applying these to the Hamiltonian for our system gives:

    qi = pim & p i = -kqi

    4) Taking the time derivative of the first equation gives:

    q..i = v i = p im

    5) Substituting in for p i gives:q..i = -km qi

    6) This gives a frequency of oscillation for each coordinate of:

    ω = km

    Printed by Wolfram Mathematica Student Edition

  • 7) These equations (one for each subscript i) have a well known solution:

    qi = Ai Sin(ωt) + Bi Cos(ωt)8) Plugging the trial solution into the equations of motion gives:

    Ai = -pi(0)mωBi = qi(0)pi = mq i

    9) Now I can use the solution above with the proper constants to determine the trajectory of the mass

    Set the problem parameters

    In[1]:= params1 = {m → 1, k → 1, ω → 1};inits1 = {x0 → RandomReal[],

    y0 → RandomReal[], px0 → RandomReal[], py0 → RandomReal[]};Equations

    In[3]:= X[t_] := -px0m ω Sin[ω t] + x0 Cos[ω t] /. params1 /. inits1;Y[t_] := -py0m ω Sin[ω t] + y0 Cos[ω t] /. params1 /. inits1;Px[t_] := -px0 Cos[ω t] - x0m ω Sin[ω t] /. params1 /. inits1;Py[t_] := -py0 Cos[ω t] - y0m ω Sin[ω t] /. params1 /. inits1;

    2 AlexWilhelm2016September22nd.nb

    Printed by Wolfram Mathematica Student Edition

  • Part A)

    In[7]:= plot1a = ListPlot[{{Labeled[{X[0], Y[0]}, "0"]}, {Labeled[{X[1], Y[1]}, "1"]},{Labeled[{X[10], Y[10]}, "10"]}}, PlotStyle → {Blue, Red, Green},PlotRange → {{-1, 1}, {-1, 1}}, AxesLabel → {"X", "Y"},PlotLabel → "Y vs X for Three Times", ImageSize → Large, Frame → True];

    Legended[plot1a, {Blue, "0 Seconds", Red, "1 Second", Green, "10 Seconds"}]

    Out[8]=

    0

    1

    10

    -1.0 -0.5 0.0 0.5 1.0-1.0

    -0.5

    0.0

    0.5

    1.0

    X

    YY vs X for Three Times

    0 Seconds

    1 Second

    10 Seconds

    AlexWilhelm2016September22nd.nb 3

    Printed by Wolfram Mathematica Student Edition

  • Part B)

    In[9]:= plot1b =ListPlot[{{Labeled[{Px[0], Py[0]}, "0"]}, {Labeled[{Px[1], Py[1]}, "1"]},{Labeled[{Px[10], Py[10]}, "10"]}}, PlotStyle → {Blue, Red, Green},AxesLabel → {"Px", "Py"}, PlotLabel → "Py vs Px for Three Times",ImageSize → Large, PlotRange → All, Frame → True];

    Legended[plot1b, {Blue, "0 Seconds", Red, "1 Second", Green, "10 Seconds"}]

    Out[10]=

    0

    1

    10

    -0.4 -0.2 0.0 0.2 0.4-0.4

    -0.2

    0.0

    0.2

    Px

    Py

    Py vs Px for Three Times

    0 Seconds

    1 Second

    10 Seconds

    4 AlexWilhelm2016September22nd.nb

    Printed by Wolfram Mathematica Student Edition

  • Part C)

    In[11]:= plot1cX = ListPlot[{{Labeled[{Px[0], X[0]}, "0"]},{Labeled[{Px[1], X[1]}, "1"]}, {Labeled[{Px[10], X[10]}, "10"]}},PlotStyle → {Blue, Red, Green}, PlotRange → All, AxesLabel → {"X", "Px"},PlotLabel → "Px vs X for Three Times", ImageSize → Large, Frame → True];

    plot1cY = ListPlot[{{Labeled[{Py[0], Y[0]}, "0"]},{Labeled[{Py[1], Y[1]}, "1"]}, {Labeled[{Py[10], Y[10]}, "10"]}},PlotStyle → {Blue, Red, Green}, PlotRange → All, AxesLabel → {"Y", "Py"},PlotLabel → "Py vs Y for Three Times", ImageSize → Large, Frame → True];

    Legended[Grid[{{plot1cX}, {plot1cY}}],{Blue, "0 Seconds", Red, "1 Second", Green, "10 Seconds"}]

    AlexWilhelm2016September22nd.nb 5

    Printed by Wolfram Mathematica Student Edition

  • Out[13]=

    0

    1

    10

    -0.4 -0.2 0.0 0.2 0.4-0.2-0.1

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    X

    Px

    Px vs X for Three Times

    0

    1

    10

    -0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3

    -0.2

    -0.1

    0.0

    0.1

    0.2

    0.3

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    Y

    Py

    Py vs Y for Three Times

    0 Seconds

    1 Second

    10 Seconds

    Problem #2Here, I consider 10 of the above systems and plot slices of their phase spaces. They are identical

    6 AlexWilhelm2016September22nd.nb

    Printed by Wolfram Mathematica Student Edition

  • systems using the same parameters but each one using it’s own random initial conditions.

    In[14]:= NN = 10;inits2 = {x0 → Table[RandomReal[], {NN}], y0 → Table[RandomReal[], {NN}],

    px0 → Table[RandomReal[], {NN}], py0 → Table[RandomReal[], {NN}]};

    X2[t_] := -px0m ω Sin[ω t] + x0 Cos[ω t] /. params1 /. inits2;Y2[t_] := -py0m ω Sin[ω t] + y0 Cos[ω t] /. params1 /. inits2;Px2[t_] := -px0 Cos[ω t] - x0m ω Sin[ω t] /. params1 /. inits2;Py2[t_] := -py0 Cos[ω t] - y0m ω Sin[ω t] /. params1 /. inits2;Xt0 = X2[0];Xt1 = X2[1];Xt10 = X2[10];Yt0 = Y2[0];Yt1 = Y2[1];Yt10 = Y2[10];Pxt0 = Px2[0];Pxt1 = Px2[1];Pxt10 = Px2[10];Pyt0 = Py2[0];Pyt1 = Py2[1];Pyt10 = Py2[10];

    AlexWilhelm2016September22nd.nb 7

    Printed by Wolfram Mathematica Student Edition

  • Part A)

    In[32]:= plot2a = ListPlot[{Thread[{Xt0, Yt0}], Thread[{Xt1, Yt1}], Thread[{Xt10, Yt10}]},PlotStyle → {Blue, Red, Green}, PlotRange → {{-1, 1}, {-1, 1}},AxesLabel → {"X", "Y"}, PlotLabel → "Y vs X for Three Times",ImageSize → Large, Frame → True];

    Legended[plot2a, {Blue, "0 Seconds", Red, "1 Second", Green, "10 Seconds"}]

    Out[33]=

    -1.0 -0.5 0.0 0.5 1.0-1.0

    -0.5

    0.0

    0.5

    1.0

    X

    YY vs X for Three Times

    0 Seconds

    1 Second

    10 Seconds

    8 AlexWilhelm2016September22nd.nb

    Printed by Wolfram Mathematica Student Edition

  • Part B)

    In[34]:= plot2b =ListPlot[{Thread[{Pxt0, Pyt0}], Thread[{Pxt1, Pyt1}], Thread[{Pxt10, Pyt10}]},PlotStyle → {Blue, Red, Green}, AxesLabel → {"Px", "Py"},PlotLabel → "Py vs Px for Three Times",ImageSize → Large, PlotRange → All, Frame → True];

    Legended[plot2b, {Blue, "0 Seconds", Red, "1 Second", Green, "10 Seconds"}]

    Out[35]=

    -1.0 -0.5 0.0 0.5 1.0-1.0

    -0.5

    0.0

    0.5

    1.0

    Px

    Py

    Py vs Px for Three Times

    0 Seconds

    1 Second

    10 Seconds

    Part C)

    In[36]:= plot2cX =ListPlot[{Thread[{Xt0, Pxt0}], Thread[{Xt1, Pxt1}], Thread[{Xt10, Pxt10}]},PlotStyle → {Blue, Red, Green}, AxesLabel → {"X", "Px"},PlotLabel → "Px vs X for Three Times",ImageSize → Large, PlotRange → All, Frame → True];

    plot2cY = ListPlot[{Thread[{Yt0, Pyt0}], Thread[{Yt1, Pyt1}],Thread[{Yt10, Pyt10}]}, PlotStyle → {Blue, Red, Green},

    AxesLabel → {"Y", "Py"}, PlotLabel → "Py vs Y for Three Times",ImageSize → Large, PlotRange → All, Frame → True];

    Legended[Grid[{{plot2cX}, {plot2cY}}],{Blue, "0 Seconds", Red, "1 Second", Green, "10 Seconds"}]

    AlexWilhelm2016September22nd.nb 9

    Printed by Wolfram Mathematica Student Edition

  • Out[38]=-0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0

    -1.0

    -0.5

    0.0

    0.5

    1.0

    X

    Px

    Px vs X for Three Times

    -0.5 0.0 0.5-1.0

    -0.5

    0.0

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    1.0

    Y

    Py

    Py vs Y for Three Times

    0 Seconds

    1 Second

    10 Seconds

    Problem #31. De Luca, a. & Scardicchio, a. Ergodicity breaking in a model showing many-body localization. EPL(Europhysics Lett. 101, 37003 (2013).

    10 AlexWilhelm2016September22nd.nb

    Printed by Wolfram Mathematica Student Edition

  • This paper discusses the fundamental assumption of ergodicity in Statistical Mechanics and its role inthe phenomena of Many Body Localization. They describe a metric called the Inverse ParticipationRatio that can be used to measure the complicated dynamics of a quantum system. The IPR is essen-tially the time-averaged probability of the time-evolution of the state ψ0. The Participation Ratio (inverseof the IPR) that is used in the paper is a strong measure of the amount of the Hilbert Space explored bythe system. By studying the PR in a 1D-type system, they are able to see emergent ergodicity-breakingphenomena arise in the presence of strong disorder. These phenomena are characterized by thelocalization of the many body wavefunction onto a small section of the Hilbert space so that the systemno longer explores all the possible configurations afforded it. Comparisons to the standard AndersonLocalization thory are made and the probability of the system returning to a previous state is heavilydiscussed. In conclusion, this paper has identified many tools and metrics in a simple quantum manybody system that will be necessary for future study of more complex many body ergodicity-breakingsystems.

    AlexWilhelm2016September22nd.nb 11

    Printed by Wolfram Mathematica Student Edition

  • Brandon Skogen

    PHGN530: Statistical Mechanics HW2

    Problem 1)

    For a Simple Harmonic Oscillator in 2D we can consider the speperability of the x and y

    coordinates with respect to the spring force and the kinetic energy of the mass, m, with spring having a

    spring constant k.

    𝐹𝑥 = 𝑚 𝑎𝑥 = 𝑚 �̈� = −𝑘 𝑥2

    𝐹𝑦 = 𝑚 𝑎𝑦 = 𝑚 �̈� = −𝑘 𝑦2

    We can then solve for x and y giving the general solutions of

    𝑥(𝑡) = 𝐶[0] 𝑐𝑜𝑠(𝜔 𝑡 + 𝛿0) + 𝐶[1] sin (𝜔 𝑡 + 𝛿1)

    𝑦(𝑡) = 𝐶[2] 𝑐𝑜𝑠(𝜔 𝑡 + 𝛿2) + 𝐶[3] sin (𝜔 𝑡 + 𝛿3)

    For 𝜔 = √𝑘/𝑚 If we set both C[1] and C[3] to zero and have the initial phase offsets

    𝛿0, 𝛿1, 𝛿2, 𝛿3 all set to zero we then have the general solution:

    𝑥(𝑡) = 𝐶[0] 𝑐𝑜𝑠(𝜔 𝑡)

    𝑦(𝑡) = 𝐶[2] 𝑐𝑜𝑠(𝜔 𝑡)

    For this case the parameters C[0] and C[2] are the max displacements for the mass with respect

    to the coordinates (0,0). To obtain the momentum we simply take a time derivative and multiply by the

    mass, m, to get:

    𝑝𝑥(𝑡) = −𝐶[0] 𝑚 𝜔 𝑠𝑖𝑛(𝜔 𝑡)

    𝑝𝑦(𝑡) = −𝐶[2] 𝑚 𝜔 𝑠𝑖𝑛(𝜔 𝑡)

    a) For a single SHO the x verses y plot for given times t=0, t=1s, t=10s are shown below

  • b) For a single SHO the px verses py plot for given times t=0, t=1s, t=10s are shown below

    c) For a single SHO the x verses px plot for given times t=0, t=1s, t=10s are shown below

  • For a single SHO the y verses py plot for given times t=0, t=1s, t=10s are shown below

    2) For multiple SHO each position is randomly chosen for x and y initial displacements the plots can be

    seen below

    a) For 10 SHOs the x verses y plot for given times t=0, t=1s, t=10s are shown below

  • b) For 10 SHOs the px verses py plot for given times t=0, t=1s, t=10s are shown below

    c) For 10 SHOs the x verses px plot for given times t=0, t=1s, t=10s are shown below

    For 10 SHOs the y verses py plot for given times t=0, t=1s, t=10s are shown below

  • It can be observed that for the x vs y and px vs py plots that each individual point moves in a linear

    motion with respect to the origin as the center of oscillation. For the x vs px and y vs py plot the points

    rotate clockwise with respect to the origin following and elliptical path determined by the initial sy

    displacement, Those with larger displacements relative to the origin have larger elliptical paths.

    3)

    Chi Ho Yeung and David Saad, “Self-sustained clusters and ergodicity breaking in spin models”

    Phys. Rev. E 88, 032132 – Published 23 September 2013

    The above research paper covers several models that suggest ergodicity breaking for self-sustained spin

    clusters. The models considered are that of Ising model for ferromagnets, nontrivial cluster formation in

    SK model spin glass. The results show that for week uniform spin coupling (𝐽0 < 1) the emergence of

    large self-sustained spin clusters contributes to reduced entropy at low temperatures (T < 0.5). The

    presence of these self-sustained spin clusters indicates metastable states which are resilient to spin

    flipping at low energies.

  • ����������������������������������������

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    �[�_] �= � * ���[� + ϕ�]��[�_] �= � * ���[� + ϕ�]�

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    4 ��� SM_HW2_HarmonicOscillators.nb

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    6 ��� SM_HW2_HarmonicOscillators.nb

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    8 ��� SM_HW2_HarmonicOscillators.nb

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    ��������[{{�[�] /� {� → �� ϕ� → π / ��}� ��[�] /� {� → �� ϕ� → π / ��}}�{�[�] /� {� → �� ϕ� → π / �}� ��[�] /� {� → �� ϕ� → π / �}}�{�[�] /� {� → �� ϕ� → π / �}� ��[�] /� {� → �� ϕ� → π / �}}�{�[�] /� {� → �� ϕ� → π / �}� ��[�] /� {� → �� ϕ� → π / �}}�{�[�] /� {� → �� ϕ� → π / �}� ��[�] /� {� → �� ϕ� → π / �}}�{�[�] /� {� → �� ϕ� → π / �}� ��[�] /� {� → �� ϕ� → π / �}}�{�[�] /� {� → �� ϕ� → π / �}� ��[�] /� {� → �� ϕ� → π / �}}�{�[�] /� {� → �� ϕ� → π / �}� ��[�] /� {� → �� ϕ� → π / �}}�{�[�] /� {� → �� ϕ� → π / �}� ��[�] /� {� → �� ϕ� → π / �}}�{�[�] /� {� → �� ϕ� → π}� ��[�] /� {� → �� ϕ� → π}}}���������� → ����� ����������� → {���������� �����}]�

    ��������[{{�[��] /� {� → �� ϕ� → π / ��}� ��[��] /� {� → �� ϕ� → π / ��}}�{�[��] /� {� → �� ϕ� → π / �}� ��[��] /� {� → �� ϕ� → π / �}}�{�[��] /� {� → �� ϕ� → π / �}� ��[��] /� {� → �� ϕ� → π / �}}�{�[��] /� {� → �� ϕ� → π / �}� ��[��] /� {� → �� ϕ� → π / �}}�{�[��] /� {� → �� ϕ� → π / �}� ��[��] /� {� → �� ϕ� → π / �}}�{�[��] /� {� → �� ϕ� → π / �}� ��[��] /� {� → �� ϕ� → π / �}}�{�[��] /� {� → �� ϕ� → π / �}� ��[��] /� {� → �� ϕ� → π / �}}�{�[��] /� {� → �� ϕ� → π / �}� ��[��] /� {� → �� ϕ� → π / �}}�{�[��] /� {� → �� ϕ� → π / �}� ��[��] /� {� → �� ϕ� → π / �}}�{�[��] /� {� → �� ϕ� → π}� ��[��] /� {� → �� ϕ� → π}}}���������� → ������ ����������� → {���������� �����}]]

    12 ��� SM_HW2_HarmonicOscillators.nb

    ����������������������������������������������

  • ��������

    ●●●●●●●●●

    ●●●●●●●●●

    ●●●●●●● ● ● ●

    ●-1.0 -0.5 0.5 1.0 y

    -1.0

    -0.5

    0.5

    1.0

    PyPy vs y

    ������������������������������������The selected literature article that shows ergodicity breaking is titled “Statistical Aging and Nonergodic-ity in the Flouresence of Single Nanocrystals” by X. Brokmann, et al. They define ergodicity as “the equivalence of the ensemble- and time-averaged mean-squared displacement”. In order to get a direct measure of ergodicity, they looked at the flouresence of 215 quatum dots in semiconductor CdSe nanostructures as a function of time. What they found was that the fluctuations of the time averaged flouresence from one QD was on the order of the time average itself, meaning that the time average for a QD does not converge for large t. Thus, no information on the amount of flouresence from the ensem-ble of quantum dots can be obtained by just looking at a single QD, therefore breaking ergodicity. In other words, to quote the source, “These data indicate ergodicity breaking: due to rare events with a duration comparable to the total acquisition time, there is no characteristic time scale over which physi-cal observables can be time averaged. Even for long acquisition time, Φ(i)on(0↦t) does not converge and no information on the ensemble value Φon can be obtained by time averaging an individual trajec-tory”, where Φon is the fraction of time the QD was in the on (or flourescing) state.URL of the article: http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.90.120601

    SM_HW2_HarmonicOscillators.nb ���13

    ����������������������������������������������

  • PHGN 530 STATISTICAL MECHANICS

    Homework 2

    Eric Jones

    September 20, 2016

    1 IMPORTANCE OF LIOUVILLE’S THEOREM IN SCIENCE (150 PTS)

    The Hamiltonian for the 2D simple harmonic oscillator with spring constant k and mass mis:

    H = p2x

    2m+

    p2y2m

    + 12

    kx2 + 12

    k y2.

    The canonical equations of motion for this system are then:

    ẋ = pxm

    ẏ = pym

    ṗx =−kx ṗy =−k y.

    Substituting the top set of equations into the bottom set, we recover Newton’s equations fora 2D simple harmonic oscillator.

    ẍ =− km

    x ÿ =− km

    y

    While the general solution is

    xi (t ) = Ai e iωx +Bi e−iωx

    1

  • withω=√

    km , we can choose initial conditions such that trajectories can be represented sim-

    ply as:

    xi (t ) = Ai cos(ωt ).

    For simplicity, we take k = m = ω = 1, note that both xi (t ) and Ai have units of length, andthat the Ai will be amplitudes randomly determined between −1 and 1. For the plot pointst = 0 s, t = 1 s, and t = 10 s the functions

    x(t ) = Ax cos(ωt ) y(t ) = Ay cos(ωt )px (t ) =−Ax sin(ωt ) py (t ) =−Ay sin(ωt )

    are plotted below.

    1.1 PART A

    1.0 0.5 0.0 0.5 1.0x

    1.0

    0.5

    0.0

    0.5

    1.0

    y

    Figure 1.1: t = 0 s

    2

  • 1.0 0.5 0.0 0.5 1.0x

    1.0

    0.5

    0.0

    0.5

    1.0

    y

    Figure 1.2: t = 1 s

    1.0 0.5 0.0 0.5 1.0x

    1.0

    0.5

    0.0

    0.5

    1.0

    y

    Figure 1.3: t = 10 s

    3

  • 1.2 PART B

    1.0 0.5 0.0 0.5 1.0px

    1.0

    0.5

    0.0

    0.5

    1.0

    py

    Figure 1.4: t = 0 s

    1.0 0.5 0.0 0.5 1.0px

    1.0

    0.5

    0.0

    0.5

    1.0

    py

    Figure 1.5: t = 1 s

    4

  • 1.0 0.5 0.0 0.5 1.0px

    1.0

    0.5

    0.0

    0.5

    1.0

    py

    Figure 1.6: t = 10 s

    1.3 PART C

    1.0 0.5 0.0 0.5 1.0xi

    1.0

    0.5

    0.0

    0.5

    1.0

    pi

    i=x

    i=y

    Figure 1.7: t = 0 s

    5

  • 1.0 0.5 0.0 0.5 1.0xi

    1.0

    0.5

    0.0

    0.5

    1.0

    pi

    i=x

    i=y

    Figure 1.8: t = 1 s

    1.0 0.5 0.0 0.5 1.0xi

    1.0

    0.5

    0.0

    0.5

    1.0

    pi

    i=x

    i=y

    Figure 1.9: t = 10 s

    6

  • 2 IMPORTANCE OF LIOUVILLE’S THEOREM IN SCIENCE (400 PTS)

    Analogous plots are produced below for ten 2D simple harmonic oscillators with randomizedapmlitudes.

    2.1 PART A

    1.0 0.5 0.0 0.5 1.0x

    1.0

    0.5

    0.0

    0.5

    1.0

    y

    Figure 2.1: t = 0 s

    7

  • 1.0 0.5 0.0 0.5 1.0x

    1.0

    0.5

    0.0

    0.5

    1.0

    y

    Figure 2.2: t = 1 s

    1.0 0.5 0.0 0.5 1.0x

    1.0

    0.5

    0.0

    0.5

    1.0

    y

    Figure 2.3: t = 10 s

    8

  • 2.2 PART B

    1.0 0.5 0.0 0.5 1.0px

    1.0

    0.5

    0.0

    0.5

    1.0

    py

    Figure 2.4: t = 0 s

    1.0 0.5 0.0 0.5 1.0px

    1.0

    0.5

    0.0

    0.5

    1.0

    py

    Figure 2.5: t = 1 s

    9

  • 1.0 0.5 0.0 0.5 1.0px

    1.0

    0.5

    0.0

    0.5

    1.0

    py

    Figure 2.6: t = 10 s

    2.3 PART C

    1.0 0.5 0.0 0.5 1.0xi

    1.0

    0.5

    0.0

    0.5

    1.0

    pi

    i=x

    i=y

    Figure 2.7: t = 0 s

    10

  • 1.0 0.5 0.0 0.5 1.0xi

    1.0

    0.5

    0.0

    0.5

    1.0

    pi

    i=x

    i=y

    Figure 2.8: t = 1 s

    1.0 0.5 0.0 0.5 1.0xi

    1.0

    0.5

    0.0

    0.5

    1.0

    pi

    i=x

    i=y

    Figure 2.9: t = 10 s

    11

  • 3 ERGODICITY (300 PTS)

    The recent (published January 22, 2015) review article by Ehud Altman and Ronen Vosk enti-tled, “Universal Dynamics and Renormalization in Many-Body-Localized Systems” presentsa few interesting perspectives on how both localized and delocalized many-body states canviolate ergodicity in condensed matter systems. The basic idea is as follows. If the dynam-ics of a many-body system are unconstrained to freely explore the Hilbert space, then theeventual state arrived at would be one of maximum entanglement entropy, correspondingto the state reached in thermal equilibrium. Hence, over long times the full Hilbert space(and hence phase sub-space consistent with macroscopic constraints) is explored and timeaverages and ensemble averages can be interchanged. If however, there is some mechanismfor many-body localization, then the time-evolution of the system is constrained to movebetween states that respect this localization. The prototypical example of this is Andersonlocalization wherein a single particle state is localized by a set of random potential wells. Interms of the renormalization group method, this can be seen as the isolation of fast-timescalefreezing out of degrees of freedom such that interactions between the single particle and theother particles (such as for example Ising model spin ensembles) can be effectively modeledas a single particle in a background potential. Thus, this many-body effect constricts the re-gion of phase space explored by each constituent body in the system and therefore also theentire system. Then, time averaged quantities of this system will not correspond to ensembleaverages of systems with variably prepared initial conditions. One experimental realizationof this effect comes in the form of phonon decoupling from electrons due to localization indisordered films at very low temperatures. In this case phonon assisted transport is frozen outand so voltage-current responses come in discrete jumps indicating that in regions betweenjumps, the states accesable to the electron population are restricted. [Ovadia M, Sacépé B,Shahar D. 2009. Phys. Rev. Lett. 102:176802]

    12

  • Homework 2#

    Consider a particle of mass m moving in the two-dimensional harmonic potential

    21( , )2

    U x y kr (1)

    Where r=2 2x y and k>0, It follows that the particle is subject to a force;

    ( , )F U k x y kr (2)

    According to Newton's second law, the equation of motion of the particle is

    2

    2

    d rm f kr

    dt (3)

    When written in component form, the above equation reduces to

    22

    02

    d xw x

    dt (4)

    22

    02

    d yw y

    dt (5)

    Where 0 /w k m and the general solutions are

    0 1cos( )x A w t (6)

    0 2cos( )y B w t (7)

    Here, A,B, 1 2 are arbitrary constants of integration. We can simplify the above equations

    slightly by shifting the origin of time (which is, after all, arbitrary): i.e., 1 0' /t t w and we

    can obtain,

    0cos( ')x A w t (8)

    0cos( ' )y A w t (9)

    Where 2 1

    In order to find out the relation between x and y, we should use trigonometry, and use equation (8)

    0 0[cos( ')cos sin( ')sin ]y B w t w t (10)

    22 2 2 2

    0 2( cos ) sin ( ')sin (1 )sin

    y x xw t

    B A A (11)

    2 22

    2 22 cos sin

    x xy y

    A AB B (12)

  • For the momentum, we have

    0 0sin( ')xP Amw w t (13)

    0 0sin( ' )yP Amw w t (14)

    We consider A=2;B=1;w=m=k=1;△=pi/4 and get the following graphs in Part I

    Part I (a)

    Figure 1 y vs x at t=0 t=1s and t=10 s

    Part I (b)

    Figure 2 Px vs t(red line) and Py vs t(blue line) at t=0 t=1s and t=10 s

  • Part I (c)

    Figure 3 Px vs x at t=0 t=1s and t=10 s

    Figure 4 Py vs y at t=0 t=1s and t=10 s

    Part II (a)

    For part II, we just need to set △ to 10 different values; the red dots represent 0s, blue ones

    represent 1s; green ones represent 10s. Here, I set △=0.3pi, 0.6pi,0.9pi....3pi

  • Figure 5 y vs x under different △ at t=0(red) t=1s(blue) and t=10s(green)

    Part II (b)

    Figure 6 Px vs t under different △ at t=0(red) t=1s(blue) and t=10s(green)

  • Figure 7 Py vs t under different △ at t=0(red) t=1s(blue) and t=10s(green)

    Part II (c)

    Figure 8 Px vs x under different △ at t=0(red) t=1s(blue) and t=10s(green)

  • Figure 9 Py vs y under different △ at t=0(red) t=1s(blue) and t=10s(green)

    Broken ergodicity reflects failure of equilibrating chemical order. In this paper (F.

    Calvo et.al Evidence for broken ergodicity due to chemical alloying from the

    dissociation kinetics of binary clusters THE JOURNAL OF CHEMICAL PHYSICS

    140, 214301 (2014) ), the authors studied the the interplay between thermal relaxation

    and statistical dissociation in binary Morse clusters using numerical simulations and

    simple statistical approaches. By varying the ranges and their mismatch, the

    simulations have shed light onto various possible behaviors at and out of thermal

    equilibrium. The short-range interaction together with the long range-interaction are

    studied and compared and they found the relaxation and dissociation times both

    increase and become closer to each other, ergodicity is restored if the cluster is

    dominated by the long-range interaction. Tracking broken ergodicity in the

    dissociation behavior of homoatomic clusters would also be possible, at least in

    principle, if they exhibit well-defined structural transitions at low temperature. I think

    this paper is very informative for me since this method can probably be used in my

    research in terms of studying the stability of different hydrate structure and different

    non-thermal stable cages.

  • Statistical Mechanics HW 2

    Kirsten Blagg

    09/22/16

    1) Simple Harmonic Oscillator

    The Hamiltonian

    Importance of Liouville’s theorem in science: Consider a simple harmonic oscillator in 2d with spring

    constant k and mass m. Plot the following:

    Looking at a 2D SHO we know that:

    V(x,y) = k

    2Ix2 + y2M

    Thus the hamiltionian is given by

    H = p

    2

    2 m+

    1

    2kq

    2

    where

    p = px x`

    + py y`

    q = x x`

    + y y`

    And we can write the Hamiltonian as

    H(x,y) = px

    2

    2 m+

    py2

    2 m+

    k

    2Ix2 + y2M

    Equations of Motion

    Using the equations of motion in Hamiltonian (keeping it written as pi and qi for simplicity)

    pi 

    =-¶H

    ¶qi= - kqi

    qi 

    =-¶H

    ¶pi=

    pi

    m

    Taking the second derivative of the position equation we get

    qi

    ..

    =pi

    m

     

    By substitution

    qi

    ..

    = -k

    mpi

    We can use the definition Ω2

    =k

    m and thus get a differential equation

    qi

    ..

    = -Ω2

    qi

    The solutions are obviously

    qi = Ai sinHΩtL + Bi cosHΩtLAnd by plugging this back into the equations of motion we get

    qi = Ai mΩ cosHΩtL - Bi mΩ sinHΩtL

    By looking at t=0 we know that

    Bi = qiH0L

    Ai =-piH0L

    mΩ

    We then get our solutions

    Printed by Wolfram Mathematica Student Edition

  • Using the equations of motion in Hamiltonian (keeping it written as pi and qi for simplicity)

    pi 

    =-¶H

    ¶qi= - kqi

    qi 

    =-¶H

    ¶pi=

    pi

    m

    Taking the second derivative of the position equation we get

    qi

    ..

    =pi

    m

     

    By substitution

    qi

    ..

    = -k

    mpi

    We can use the definition Ω2

    =k

    m and thus get a differential equation

    qi

    ..

    = -Ω2

    qi

    The solutions are obviously

    qi = Ai sinHΩtL + Bi cosHΩtLAnd by plugging this back into the equations of motion we get

    qi = Ai mΩ cosHΩtL - Bi mΩ sinHΩtL

    By looking at t=0 we know that

    Bi = qiH0L

    Ai =-piH0L

    mΩ

    We then get our solutions

    In[2]:= x@px00_, x00_, t_D = -px00 Hm * ΩL * Sin@Ω * tD + x00 * Cos@Ω * tD;y@py00_, y00_, t_D = -py00 Hm * ΩL * Sin@Ω * tD + y00 * Cos@Ω * tD;px@px00_, x00_, t_D = -px00 * Cos@Ω * tD - x00 Hm * ΩL * Sin@Ω * tD;py@py00_, y00_, t_D = -py00 * Cos@Ω * tD - y00 Hm * ΩL * Sin@Ω * tD;

    Determination of parameters

    For simplicity we will set m=k=1 so that Ω = 1 and will randomly generate initial conditions

    In[6]:= param = 8m ® 1, k ® 1, Ω ® 1

  • In[16]:= ListPlot@8Labeled@8x@px0, x0, 0D, y@py0, y0, 0D< . param, "0"D,Labeled@8x@px0, x0, 1D, y@py0, y0, 1D< . param, "1"D,Labeled@8x@px0, x0, 10D, y@py0, y0, 10D< . param, "10"D

  • In[17]:= ListPlot@8Labeled@8px@px0, x0, 0D, x@px0, x0, 0D< . param, "0"D,Labeled@8px@px0, x0, 1D, x@px0, x0, 1D< . param, "1"D,Labeled@8px@px0, x0, 10D, x@px0, x0, 10D< . param, "10"D

  • In[32]:= x010 = Table@RandomReal@D, 810

  • In[49]:= GraphicsRow@8ListPlot@Table@8px10t0@@nDD, py10t0@@nDD

  • Borgonovi, F., et al. “Broken ergodicity in classically chaotic spin systems.” Journal of statistical physics

    116.5-6 (2004): 1435-1447.

    The paper examined a simulation of a one dimensional classically chaotic spin chain with asymetric

    coupling. They looked at two systems of coupling: nearest neighbor and all to all. They found that the all

    to all interactions the motion did not cover the entire energy surface. Using this topology they were able

    to define a energy threshold at which some energy regions are not accessable, a place below which

    ergodicity is broken. They found this energy therhold for non-ergodicity held for any system of finite

    particles. Interestingly they found the in the nearest neighbor interactions there was no energy threshold

    and that all of the energy space was occupied. However, they could not conclude that this leads to

    ergodicity. It was most interesting that even in the presence of choas (positive Lyponov exponent) there

    were still energy regions which could not be occupied indication a non-erogdic system.

    statmech_2.nb 7

    Printed by Wolfram Mathematica Student Edition

  • HW2Marie McLain22 September 2016Collaborator(s): Kirsten Blagg, (Andrew Proudian)

    Clear"Global`*"1.) Importance of Liouville’s theorem in science. Consider a simple harmonic oscillator in 2D with spring constant k, mass m.

    The Hamiltonian of a SHO in 2D, with ω = km

    , is

    H[p_, q_] := p2 m2

    + 1 / 2 m ω2 q2H[x_, y_, px_, py_] := px2 + py2

    2 m+ 12m ω2 x2 + y2

    Using Hamilton’s equations of motion, p i = -∂qiH = -m ω2 qiq i = ∂qiH = pimtaking another derivative and substituting, we find the general solutionm q

    ..i= p

    i

    q..

    i= - ω2 qi qi = Ai Cos[ωt] + Bi Sin[ωt] for which we can substitute back into Hamilton’s equations to find momentum. pi = -mωAi Sin[ωt] + mω Cos[ωt] Define parameters in terms of initial conditions and re-write solutions. qi(0) = Ai pi = -mωBi qi = qi(0)Cos[ωt] - pi (0)

    m ω Sin[ωt] pi = -mω qi(0)Sin[ωt] + pi(0)Cos[ωt]Set constants.

    In[20]:= ω = 1; m = 1;In[21]:= x[t_] := x0 Cos[ω t] - px0 Sin[ω t]

    px[t_] := -m ω x0 Sin[ω t] + px0 Cos[ω t]y[t_] := y0 Cos[ω t ] - py0 Sin[ω t]py[t_] := -m ω y0 Sin[ω t] + py0 Cos[ω t]Generate initial conditions.

    In[25]:= x0 = RandomReal[{0.4, 0.6}];px0 = RandomReal[{0.4, 0.6}];y0 = RandomReal[{0.4, 0.6}];py0 = RandomReal[{0.4, 0.6}];Plot:

    Printed by Wolfram Mathematica Student Edition

  • a.) y vs x at t=0, t=1s, and t=10s.

    In[29]:= ListPlot{{x[0], y[0]}}, PlotLabel → "y vs. x, t=0", AxesLabel → {"x", "y"}

    Out[29]=

    0.2 0.4 0.6 0.8 1.0x

    0.2

    0.4

    0.6

    0.8

    1.0

    yy vs. x, t=0

    In[30]:= ListPlot{{x[1], y[1]}}, PlotLabel → "y vs. x, t=1", AxesLabel → {"x", "y"}

    Out[30]=

    -0.20 -0.15 -0.10 -0.05 x

    -0.25-0.20-0.15-0.10-0.05

    yy vs. x, t=1

    2 MarieMcLain2016September22.nb

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  • ListPlot{{x[10], y[10]}}, PlotLabel → "y vs. x, t=10", AxesLabel → {"x", "y"}

    Out[31]=

    -0.35 -0.30 -0.25 -0.20 -0.15 -0.10 -0.05 x

    -0.35-0.30-0.25-0.20-0.15-0.10-0.05

    yx vs. y, t=10

    b.) py vs. px at t=0, t=1s, and t=10s.

    In[33]:= ListPlot{{px[0], py[0]}}, PlotLabel → "py vs. px, t=0", AxesLabel → {"px", "py"}

    Out[33]=

    0.2 0.4 0.6 0.8px

    0.2

    0.4

    0.6

    0.8

    1.0

    pypy vs. px, t=0

    In[34]:= ListPlot{{px[1], py[1]}}, PlotLabel → "py vs. px, t=1", AxesLabel → {"px", "py"}

    Out[34]=

    -0.35 -0.30 -0.25 -0.20 -0.15 -0.10 -0.05 px

    -0.35-0.30-0.25-0.20-0.15-0.10-0.05

    pypy vs. px, t=1

    MarieMcLain2016September22.nb 3

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  • In[35]:= ListPlot{{px[10], py[10]}}, PlotLabel → "py vs. px, t=10", AxesLabel → {"px", "py"}

    Out[35]=

    -0.20 -0.15 -0.10 -0.05 px

    -0.25-0.20-0.15-0.10-0.05

    pypy vs. px, t=10

    c.) px vs x and py vs yat t=0, t=1s, t=10s

    In[36]:= ListPlot{{px[0], x[0]}}, PlotLabel → "px vs. x, t=0", AxesLabel → {"px", "x"}

    Out[36]=

    0.2 0.4 0.6 0.8px

    0.2

    0.4

    0.6

    0.8

    1.0

    xpx vs. x, t=0

    In[37]:= ListPlot{{px[0], x[0]}}, PlotLabel → "py vs. y, t=0", AxesLabel → {"py", "y"}

    Out[37]=

    0.2 0.4 0.6 0.8py

    0.2

    0.4

    0.6

    0.8

    1.0

    ypy vs. y, t=0

    4 MarieMcLain2016September22.nb

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  • In[38]:= ListPlot{{px[1], x[1]}}, PlotLabel → "px vs. x, t=1", AxesLabel → {"px", "x"}

    Out[38]=

    -0.35 -0.30 -0.25 -0.20 -0.15 -0.10 -0.05 px

    -0.20-0.15-0.10-0.05

    xpx vs. x, t=1

    In[39]:= ListPlot{{px[1], x[1]}}, PlotLabel → "py vs. y, t=1", AxesLabel → {"py", "y"}

    Out[39]=

    -0.35 -0.30 -0.25 -0.20 -0.15 -0.10 -0.05 py

    -0.20-0.15-0.10-0.05

    ypy vs. y, t=1

    In[40]:= ListPlot{{px[10], x[10]}}, PlotLabel → "px vs. x, t=10", AxesLabel → {"px", "x"}

    Out[40]=

    -0.20 -0.15 -0.10 -0.05 px

    -0.35-0.30-0.25-0.20-0.15-0.10-0.05

    xpx vs. x, t=10

    MarieMcLain2016September22.nb 5

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  • In[41]:= ListPlot{{px[10], x[10]}}, PlotLabel → "py vs. y, t=10", AxesLabel → {"py", "y"}

    Out[41]=

    -0.20 -0.15 -0.10 -0.05 py

    -0.35-0.30-0.25-0.20-0.15-0.10-0.05

    ypy vs. y, t=10

    2.) Considering 10 simple harmonic oscillators

    In[42]:= Clear[x0, px0, y0, py0]I establish initial conditions for each

    In[45]:= x0 = RandomReal[{0.4, 0.6}, 10];px0 = RandomReal[{0.4, 0.6}, 10];y0 = RandomReal[{0.4, 0.6}, 10];py0 = RandomReal[{0.4, 0.6}, 10];Plot:a.) y vs x at t=0, t=1s, and t=10s.

    In[54]:= ShowTableListPlotx[0]i, y[0]i,PlotLabel → "y vs. x, t=0", AxesLabel → {"x", "y"}, i, 10

    Out[54]=

    0.2 0.4 0.6 0.8 1.0x

    0.2

    0.4

    0.6

    0.8

    1.0

    yy vs. x, t=0

    6 MarieMcLain2016September22.nb

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  • In[55]:= ShowTableListPlotx[1]i, y[1]i,PlotLabel → "y vs. x, t=1", AxesLabel → {"x", "y"}, i, 10

    Out[55]=

    -0.20 -0.15 -0.10 -0.05 x

    -0.15

    -0.10

    -0.05

    yy vs. x, t=1

    In[56]:= ShowTableListPlotx[10]i, y[10]i,PlotLabel → "y vs. x, t=10", AxesLabel → {"x", "y"}, i, 10

    Out[56]=

    -0.35 -0.30 -0.25 -0.20 -0.15 -0.10 -0.05 x

    -0.4

    -0.3

    -0.2

    -0.1

    yy vs. x, t=10

    b.) py vs. px at t=0, t=1s, and t=10s.

    MarieMcLain2016September22.nb 7

    Printed by Wolfram Mathematica Student Edition

  • In[58]:= ShowTableListPlotpx[0]i, py[0]i,PlotLabel → "py vs. px, t=0", AxesLabel → {"px", "py"}, i, 10

    Out[58]=

    0.2 0.4 0.6 0.8px

    0.2

    0.4

    0.6

    0.8

    pypy vs. px, t=0

    In[59]:= ShowTableListPlotpx[1]i, py[1]i,PlotLabel → "py vs. px, t=1", AxesLabel → {"px", "py"}, i, 10

    Out[59]=

    -0.35 -0.30 -0.25 -0.20 -0.15 -0.10 -0.05 px

    -0.4-0.3-0.2-0.1

    pypy vs. px, t=1

    8 MarieMcLain2016September22.nb

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  • In[60]:= ShowTableListPlotpx[10]i, py[10]i,PlotLabel → "py vs. px, t=10", AxesLabel → {"px", "py"}, i, 10

    Out[60]=

    -0.20 -0.15 -0.10 -0.05 px

    -0.15

    -0.10

    -0.05

    pypy vs. px, t=10

    c.) px vs x and py vs yat t=0, t=1s, t=10s

    In[64]:= ShowTableListPlotpx[0]i, x[0]i,PlotLabel → "px vs. x, t=0", AxesLabel → {"x", "px"}, i, 10

    Out[64]=

    0.2 0.4 0.6 0.8x

    0.2

    0.4

    0.6

    0.8

    1.0

    pxpx vs. x, t=0

    MarieMcLain2016September22.nb 9

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  • In[65]:= ShowTableListPlotpy[0]i, y[0]i,PlotLabel → "py vs. y, t=0", AxesLabel → {"y", "py"}, i, 10

    Out[65]=

    0.2 0.4 0.6 0.8y

    0.2

    0.4

    0.6

    0.8

    1.0

    pypy vs. y, t=0

    In[63]:= ShowTableListPlotpx[1]i, x[1]i,PlotLabel → "px vs. x, t=0", AxesLabel → {"x", "px"}, i, 10

    Out[63]=

    -0.35 -0.30 -0.25 -0.20 -0.15 -0.10 -0.05 x

    -0.25-0.20-0.15-0.10-0.05

    pxpx vs. x, t=0

    10 MarieMcLain2016September22.nb

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  • In[66]:= ShowTableListPlotpy[1]i, y[1]i,PlotLabel → "py vs. y, t=1", AxesLabel → {"y", "py"}, i, 10

    Out[66]=

    -0.4 -0.3 -0.2 -0.1 y

    -0.15

    -0.10

    -0.05

    pypy vs. y, t=1

    In[67]:= ShowTableListPlotpx[10]i, x[10]i,PlotLabel → "px vs. x, t=10", AxesLabel → {"x", "px"}, i, 10

    Out[67]=

    -0.20 -0.15 -0.10 -0.05 x

    -0.35-0.30-0.25-0.20-0.15-0.10-0.05

    pxpx vs. x, t=10

    MarieMcLain2016September22.nb 11

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  • In[68]:= ShowTableListPlotpy[10]i, y[10]i,PlotLabel → "py vs. y, t=10", AxesLabel → {"y", "py"}, i, 10

    Out[68]=

    -0.15 -0.10 -0.05 y

    -0.4

    -0.3

    -0.2

    -0.1

    pypy vs. y, t=10

    The conclusion I draw from the first two problems demonstrates the Liouville theorem. The result from part 1.) leads to the results from part 2.), meaning that the location of a solution in phase space indi-cates that those solutions are localized to a particular area of the phase space. Because we are only plotting 2D maps of the 4D system, we are just looking at slices of the higher dimensional phase space. Looking at 10 SHOs, we see occupation of the same area (volume) of the phase space.

    3.) Ergodicity -- Breakdown of ergodicity

    In the paper “Ergodicity breaking in random media and the foundation of fractional kinetics” by G. Pagnini et. al. describe a modeling framework to describe the nonergodic behavior of some biological systems. They explore diffusion in the case when a Gaussian process has long-range correlations, meaning that they have a random complex medium. This breaks a critical assumption, that of a trapping mechanism, that is usually taken by a continuous time random walk simulation. This modeling platform has been compared to the underlying stochastic process that describes motion of individual mRNA and even live E. coli cells. The important point of ergodicity stems from the random quantity in the random medium, meaning that all of the microstates are NOT equiprobable over long periods of time, as is the case in the ergodicity hypothesis. http://arxiv.org/pdf/1508.01361.pdf

    12 MarieMcLain2016September22.nb

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    Homework 2 Due 09202016AlexWilhelm2016September22ndBrandonSkogen2016September21DavidHensle_HW2EricJones2016September22HW2 Shuo LiuKirstenBlagg2016September22MarieMcLain2016September22