Infinite Square Well Wavefunction - Quantum Mechanics - Solved Past Paper, Exams of Physics

Download Infinite Square Well Wavefunction - Quantum Mechanics - Solved Past Paper and more Exams Physics in PDF only on Docsity! Physics 137A — Quantum Mechanics — Fall 2012 Midterm I - Solutions 1 Infinite Square Well Wavefunction Consider the infinite square well of width a (extending from x = 0 to x = a). Take the initial spatial wave function to be a parabola centered around the middle of the well: Ψ(x, 0) = A(ax− x2) Ψ(x, 0) x x = a where A is some constant, a is the width of the well, and where this function applies only inside the well (outside the well, Ψ(x, 0) = 0). (a) Normalize this wavefunction. (b) Calculate the expectation values 〈p〉 and 〈p2〉 in this state. (c) Given that 〈x〉 = a/2 and 〈x2〉 = 2a2/7, what are the uncertainties σx and σp? Is the uncertainty principle satisfied? (d) We can write this wavefunction as a sum over the energy eigenfunctions: Ψ(x, 0) = ∞∑ i=1 cnψn(x) where cn are some constants and ψn(x) are the energy eigenfunctions. Set up a calculation of the coefficients, cn, and write it as an integral with everything spelled out explicitly; you do not have to evaluate the integral. Are all of the cn non-zero? Write a short argument justifying your answer. Solution (a) The normalization condition for this wavefunction is the following 1 = ∫ a 0 Ψ∗(x, 0)Ψ(x, 0)dx Plugging in our wavefunction gives us 1 = |A|2 ∫ a 0 (ax− x2)2dx = |A|2 ∫ a 0 (a2x2 − 2ax3 + x4)dx. These are straightforward integrations to perform so we find 1 = |A|2 ( a2x3 3 − 2ax 4 4 + x5 5 )∣∣∣∣a 0 = |A|2a5 ( 1 3 − 1 2 + 1 5 ) = |A|2 a 5 30 . This means that we should take A = √ 30 a5 . 1 Note that we could have multiplied this by a phase factor, but we chose to have a real value for A. (b) The expectation value of p is given by the following: 〈p〉 = ∫ a 0 Ψ∗(x, 0)p̂Ψ(x, 0)dx. This is a case where we can jump to the answer. The wavefunction itself is even with respect to the center of the well. One derivative of this even function produces an odd function. So the overall integrand is the product of an even function with an odd function, which is an odd function. Any odd function integrated over a symmetric interval is zero. So 〈p〉 = 0. To find 〈p2〉 we have to compute 〈 p2 〉 = ∫ a 0 Ψ∗(x, 0)p̂p̂Ψ(x, 0)dx. As part of this consider the action of p̂2 = −~2∂2/∂x2. When acted on the wavefunction, we find p̂2Ψ(x, 0) = −~2 ∂ 2 ∂x2 A(ax− x2) = −~2A(−2) = 2A~2. So our expectation value is just 〈 p2 〉 = |A|2 ∫ a 0 (ax− x2) · 2~2dx = 2~2 30 a5 ( ax2 2 − x 3 3 )∣∣∣∣a 0 = 2~2 30 a5 a3 ( 1 2 − 1 3 ) = 2~2 30 a2 1 6 = 10~2 a2 . (c) The dispersion in x is given by σx = √ 〈x2〉 − 〈x〉2 = √ 2a2 7 − a 2 4 = a √ 2/7− 1/4 = a √ 1/28. A similar calculation for p gives σp = √ 〈p2〉 − 〈p〉2 = √ 10~2/a2 = ~ a √ 10. Their product is σxσp = a√ 28 ~ a √ 10 = ~ 2 √ 10 7 So the uncertainty principle is satisfied because √ 10/7 ≥ 1. (d) To find the coefficients cn one need only take the integral of the nth energy eigenfunction against the initial conditions: 2 The same is true when the initial conditions are a sum of energy eigenfunctions because the Schrödinger Equation is linear. So the initial conditions for this problem can become time dependent by adding the exponential factors appropriately: Ψ(x, t) = 1 2 β1(x)e −iE1t/~ + 1− i 2 β2(x)e −iE2t/~ + 1 2 β3(x)e −iE3t/~. Next we apply the definition of the expectation value of the energy: 〈H〉 = ∫ Ψ∗(x, t)ĤΨ(x, t)dx. However, we can simplify our life by realizing that our wavefunction is written as a sum of the energy eigenfunctions. So we have ĤΨ(x, t) = Ĥ ∑ i ciβi(x)e −iEit/~ = ∑ i ciEiβi(x)e −iEit/~. Now, when we multiply this by Ψ∗, and integrate, we will find again nine terms: three of which are products of an energy eigenfunction with itself, and six of which are cross terms. In each of the terms, the time dependence drops out, because each copy of exp (−iEit/~) is multiplied by exp (iEit/~), which is 1. The cross terms cancel because of the orthogonality so we are left with 〈H〉 = ∑ i |ci|2Ei ∫ β∗i (x)βi(x)dx. In the previous we have pulled the constants out of the integral. What is left is the normalization condition for the energy eigenfunctions. 〈H〉 = ∑ i |ci|2Ei This is an equally valid starting location. Notice that there is no time dependence to the expectation value of the energy. This is because we are working with stationary states. All that is left is to plug in our particular values and find the expectation value: 〈H〉 = ∣∣∣∣12 ∣∣∣∣ 1 eV + ∣∣∣∣1− i2 ∣∣∣∣ 2 eV + ∣∣∣∣12 ∣∣∣∣ 3 eV = 1 eV4 + 2 eV2 + 3 eV4 = 2 eV. Again, the reason this expectation value is not time dependent is because the energy eigenfunc- tions are able to easily furnish a solution of the full Schrödinger Equation. These solutions, which are orthogonal and normalized, cannot give time dependent energy expectation values because the time dependence always meets its conjugate and becomes a factor of 1. This is why these are called stationary states: once they are in such a state, they stay there (up to a phase factor). (c) Normally you would not expect the expectation value of the energy to be equal to one of the possible values you might get upon measuring the energy. The possible results of a measurement of energy are the energy eigenvalues. In this case, we do have an energy eigenvalue equal to 2 eV. From postulate three, we know that the probability of getting this result is given by the square of the expansion coefficient: P (Ei) = |ci|2 = | 〈βi|Ψ〉 |2. In this case, we are looking for the case when i = 2, so we take either P (2 eV) = ∣∣∣∣1− i2 ∣∣∣∣2 = 12 or we take the norm squared of 〈β2 |Ψ〉 = ∫ β∗2(x)Ψ(x, 0)dx = 3∑ i=1 ci ∫ β∗2(x)βi(x)dx = 3∑ i=1 ciδ2,i = c2 = 1− i 2 5 which is the same as the previous probability. Again, this is not generally possible for a given system and given initial conditions. The expec- tation value of the energy could be a value of energy that would never be the result of a measurement. (d) To measure the expectation value of the energy, we would prepare many copies of the system in the prescribed initial conditions. Then, the energy of each copy of the system is measured, once for each copy of the system. We can then take the average of the measured values on this collection of systems, and that would be the expectation value of the energy. 6