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camarasi
2017-06-10 03:29:33
Definition of expectation value of an operator O is\r\n\r\n< O > = <\\psi|O\\psi>\r\n\r\nPlugging in our state \\psi...\r\n\r\n<\\psi|O\\psi> = <\\frac{1}{\\sqrt{6}} \\psi_{-1}+\\frac{1}{\\sqrt{2}}\\psi_{1}+\\frac{1}{\\sqrt{3}}\\psi_{2}|O(\\frac{1}{\\sqrt{6}} \\psi_{-1}+\\frac{1}{\\sqrt{2}}\\psi_{1}+\\frac{1}{\\sqrt{3}}\\psi_{2})>\r\n\r\nNow we \'multiply out\' the inner product. This is similar to squaring a term like (a+b+c). You will get 9 terms - messy!\r\n\r\nBut remember that \r\n\r\n<\\psi_j|\\psi_k> = \\delta_{jk}\r\n\r\nwhere \\delta_{jk} is the Kronecker delta function.\r\n\r\nSo only 3 terms survive.\r\n\r\n<\\psi|O\\psi> = <\\frac{1}{\\sqrt{6}}\\psi_{-1}|O\\frac{1}{\\sqrt{6}}\\psi_{-1}> + <\\frac{1}{\\sqrt{2}}\\psi_1|O\\frac{1}{\\sqrt{2}}\\psi_1> + <\\frac{1}{\\sqrt{3}}\\psi_2|O\\frac{1}{\\sqrt{3}}\\psi_2>\r\n\r\n=\\frac{1}{6}<\\psi_{-1}|O\\psi_{-1}> + \\frac{1}{2}<\\psi_{1}|O\\psi_{1}> + \\frac{1}{3}<\\psi_{2}|O\\psi_{2}>\r\n\r\nProblem tells us that \r\n\r\n<\\psi_{-1}|O\\psi_{-1}> = -1\r\n<\\psi_{1}|O\\psi_{1}> = 1\r\n<\\psi_{2}|O\\psi_{2}> = 2\r\n\r\nUsing these, \r\n\r\n<\\psi|O\\psi> = -\\frac{1}{6} + \\frac{1}{2} + \\frac{2}{3} = 1\r\n\r\nAnswer: (C)\r\n
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camarasi
2017-06-10 03:07:40
Definition of expectation value of an operator O is \r\n\r\n \\eq <\\psi|O\\psi>\r\n\r\n
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IvanHin
2017-05-01 04:18:12
it seems answer A also satisfies your test.\r\n
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rweads
2017-04-30 01:28:20
since the liquid does not boil, and the heat element has been in the water for a long time, the system must be in thermal equilibrium where the rate of heat added is equal to the rate that heat is dissipated. It is then very reasonable that for small deviations from that equilibrium temperature the rate of heat loss will be the same.
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OptimusPrime
2017-04-08 04:47:54
This is really helpful! The point of fastest velocity of the rope shaking happens when crossing the y-axis, which is maximum E amplitude in this case.
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OptimusPrime
2017-04-08 02:27:42
(A) and (B) are maybes so far.\r\n(C) - No, we are dealing with PROTONS instead of the usual electrons. So the mass of the electron shouldn\'t come into play at all.\r\n(D) - Maybe.\r\n(E) - No, since after this collision the photon must lose energy. Recall that higher frequency means higher energy. So if we are losing energy, that means our frequency has decreased, and since frequency and wavelength are inversely related, the wavelength has increased. \r\n\r\nIf you remember the Compton equation then you\'ll find as Yosun wrote that change in wavelength equals the Compton wavelength. But, where can we extract any numbers from that?! We can\'t, so (A) and (B) are out.\r\n\r\nThat leaves (D). On top of that, (D) has the correct dimensions. [h/mc] = (M*(L^2) / T) / (M*L / T) = L. Length as in wavelength.
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OptimusPrime
2017-04-08 00:55:51
I\'m confused. Choice (E) resembles the function y = \\frac{1}{x^2} which has positive concavity, not negative like the graph shows.
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OptimusPrime
2017-04-08 00:36:49
For (B), plugging in \\\\\\\\theta = 90 degrees gives 90/2 = 45 degrees. Then, cos(45) = \\\\\\\\frac{\\\\\\\\sqrt{2}}{2}. It also has the correct dimensions. How else do we eliminate (B)?
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liuyuhang599
2017-04-04 14:43:16
Your answer is correct, \\\\\\\\1s^2 gives two possible states ( |+-> - |-+>) the singlet state, and ( |+-> +|-+> ) the triplet state. And because electron is fermion, only the antisymmetric singlet state is allowed.
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godkun
2017-04-03 19:31:06
你是傻逼吗?儿子?
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dipanshugupta
2017-03-31 10:12:40
Here \\\'s my method, using \\makebf{Picking Numbers}. Pick a time, say 12, because it is divisible both by 24 and 36. In 12 mins, \\alpha decays \\frac{1}{4} and \\beta decays \\frac{1}{6} . Add them up to get \\frac{5}{12} , which is close to half. Take a time little more than 12. Hence, 14.4 minutes. Pick C.
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dipanshugupta
2017-03-30 17:10:35
भाई चाइनीज़ मैं क्युन बात कर रहे हो?
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dipanshugupta
2017-03-30 07:35:09
If you have ever studied the Harmonic Oscillator, you\'ll know that a and a^+ are Creation and Annihilation operators, i.e, they increase or decrease one quanta of Energy. \r\nI. The commutator gives one negative quanta of energy. \r\nII. Wrong if III is correct. \r\nIII. Definitely, as per above discussion. \r\n\r\nThus, answer is C. \r\n\r\n(I have taken a QFT class so it was easier for me but it\'s in QM too).
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dipanshugupta
2017-03-29 12:28:54
Really good answer. Thank you!
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dipanshugupta
2017-03-29 10:45:02
How do you get the B cos(\\omega t) term for B.dA?
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dipanshugupta
2017-03-29 10:02:01
Simpler answer. Beta Decay is a Weak Interaction. Weak Interactions are Parity Violating. Parity is a Reflection invariance as P^\' (\\vec{x},t)  \\Rightarrow P ( - \\vec{x},t). Hence, problem solved.
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dipanshugupta
2017-03-29 08:55:09
WoolfianOperator, you used v_2 as 0.6v when you should\\\\\\\'ve used it as -0.6v, since the He atom is travelling in the opposite direction after collision.
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NervousWreck
2017-03-28 17:21:16
The most straitforward is to draw a PV diagram where an adiabatic expansion would be a curve, which is steeper than an isoterm. Therefore one immediately notices that the E is False.
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NervousWreck
2017-03-28 11:00:04
also disregard people trying to find a mistake in this solution. this is the fastest one.
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NervousWreck
2017-03-28 10:25:53
Same mass as for the nuclear warhead man, same mass.
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NervousWreck
2017-03-28 09:50:19
Yes very good man thanks.
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NervousWreck
2017-03-28 09:45:40
This is a very complicated problem due to the lack of time. However the fasters solution as I see it is the following\r\n1/(\\sqrt N/2)=0.02. Here rate is 2, which is multiplied by 1%. On the LHS is an error of poissson distribution with 2 measurements per second taken into account. The solution gives N = 5000, which now corresponds to seconds.
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NervousWreck
2017-03-28 09:21:55
From some accurate integration for a diffraction from a circular aperture
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NervousWreck
2017-03-28 09:12:16
This is a very good solution and I used it too.
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blabla0001
2017-03-25 09:26:30
this solution is too good for me. it is also so good that it becomes dangerous. hope i ll never need it again
 
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