Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. I do not see how, based on the inelastic tunneling experiments, one can still have doubts that the particle did, in fact, physically traveled through the barrier, rather than simply appearing at the other side. But for the quantum oscillator, there is always a nonzero probability of finding the point in a classically forbidden region; in other words, there is a nonzero tunneling probability. Although the potential outside of the well is due to electric repulsion, which has the 1/r dependence shown below. Note the solutions have the property that there is some probability of finding the particle in classically forbidden regions, that is, the particle penetrates into the walls. Turning point is twice off radius be four one s state The probability that electron is it classical forward A region is probability p are greater than to wait Toby equal toe. /D [5 0 R /XYZ 125.672 698.868 null] \int_{\sqrt{2n+1} }^{+\infty }e^{-y^{2}}H^{2}_{n}(x) dy, (4.298). Which of the following is true about a quantum harmonic oscillator? Energy eigenstates are therefore called stationary states . We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. I'm not so sure about my reasoning about the last part could someone clarify? Unfortunately, it is resolving to an IP address that is creating a conflict within Cloudflare's system. [3] P. W. Atkins, J. de Paula, and R. S. Friedman, Quanta, Matter and Change: A Molecular Approach to Physical Chemistry, New York: Oxford University Press, 2009 p. 66. Hmmm, why does that imply that I don't have to do the integral ? Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. /Border[0 0 1]/H/I/C[0 1 1] He killed by foot on simplifying. $\psi \left( x,\,t \right)=\frac{1}{2}\left( \sqrt{3}i{{\phi }_{1}}\left( x \right){{e}^{-i{{E}_{1}}t/\hbar }}+{{\phi }_{3}}\left( x \right){{e}^{-i{{E}_{3}}t/\hbar }} \right)$. Published since 1866 continuously, Lehigh University course catalogs contain academic announcements, course descriptions, register of names of the instructors and administrators; information on buildings and grounds, and Lehigh history. >> This distance, called the penetration depth, \(\delta\), is given by The turning points are thus given by En - V = 0. Classically, the particle is reflected by the barrier -Regions II and III would be forbidden According to quantum mechanics, all regions are accessible to the particle -The probability of the particle being in a classically forbidden region is low, but not zero -Amplitude of the wave is reduced in the barrier MUJ 11 11 AN INTERPRETATION OF QUANTUM MECHANICS A particle limited to the x axis has the wavefunction Q. Lehigh Course Catalog (1999-2000) Date Created . "Quantum Harmonic Oscillator Tunneling into Classically Forbidden Regions" Quantum mechanically, there exist states (any n > 0) for which there are locations x, where the probability of finding the particle is zero, and that these locations separate regions of high probability! Peter, if a particle can be in a classically forbidden region (by your own admission) why can't we measure/detect it there? Using indicator constraint with two variables. Last Post; Nov 19, 2021; /D [5 0 R /XYZ 261.164 372.8 null] Although it presents the main ideas of quantum theory essentially in nonmathematical terms, it . where is a Hermite polynomial. /Length 1178 (vtq%xlv-m:'yQp|W{G~ch iHOf>Gd*Pv|*lJHne;(-:8!4mP!.G6stlMt6l\mSk!^5@~m&D]DkH[*. In particular the square of the wavefunction tells you the probability of finding the particle as a function of position. The same applies to quantum tunneling. \[T \approx e^{-x/\delta}\], For this example, the probability that the proton can pass through the barrier is . You are using an out of date browser. represents a single particle then 2 called the probability density is the from PHY 1051 at Manipal Institute of Technology It only takes a minute to sign up. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. +2qw-\ \_w"P)Wa:tNUutkS6DXq}a:jk cv It may not display this or other websites correctly. This shows that the probability decreases as n increases, so it would be very small for very large values of n. It is therefore unlikely to find the particle in the classically forbidden region when the particle is in a very highly excited state. \int_{\sqrt{7} }^{\infty }(8y^{3}-12y)^{2}e^{-y^{2}}dy=3.6363. \int_{\sqrt{3} }^{\infty }y^{2}e^{-y^{2}}dy=0.0495. Particle in a box: Finding <T> of an electron given a wave function. A particle absolutely can be in the classically forbidden region. Probability of particle being in the classically forbidden region for the simple harmonic oscillator: a. ~ a : Since the energy of the ground state is known, this argument can be simplified. In fact, in the case of the ground state (i.e., the lowest energy symmetric state) it is possible to demonstrate that the probability of a measurement finding the particle outside the . Related terms: Classical Approach (Part - 2) - Probability, Math; Video | 09:06 min. This superb text by David Bohm, formerly Princeton University and Emeritus Professor of Theoretical Physics at Birkbeck College, University of London, provides a formulation of the quantum theory in terms of qualitative and imaginative concepts that have evolved outside and beyond classical theory. Thanks for contributing an answer to Physics Stack Exchange! Third, the probability density distributions for a quantum oscillator in the ground low-energy state, , is largest at the middle of the well . I view the lectures from iTunesU which does not provide me with a URL. Disconnect between goals and daily tasksIs it me, or the industry? /Filter /FlateDecode \int_{\sqrt{5} }^{\infty }(4y^{2}-2)^{2} e^{-y^{2}}dy=0.6740. For a quantum oscillator, assuming units in which Planck's constant , the possible values of energy are no longer a continuum but are quantized with the possible values: . Mesoscopic and microscopic dipole clusters: Structure and phase transitions A.I. When the tip is sufficiently close to the surface, electrons sometimes tunnel through from the surface to the conducting tip creating a measurable current. Have you? The connection of the two functions means that a particle starting out in the well on the left side has a finite probability of tunneling through the barrier and being found on the right side even though the energy of the particle is less than the barrier height. Go through the barrier . classically forbidden region: Tunneling . Correct answer is '0.18'. That's interesting. Ok. Kind of strange question, but I think I know what you mean :) Thank you very much. Mutually exclusive execution using std::atomic? khloe kardashian hidden hills house address Danh mc I asked my instructor and he said, "I don't think you should think of total energy as kinetic energy plus potential when dealing with quantum.". Why Do Dispensaries Scan Id Nevada, >> >> %PDF-1.5 You'll get a detailed solution from a subject matter expert that helps you learn core concepts. In general, we will also need a propagation factors for forbidden regions. In this approximation of nuclear fusion, an incoming proton can tunnel into a pre-existing nuclear well. And I can't say anything about KE since localization of the wave function introduces uncertainty for momentum. A few that pop in my mind right now are: Particles tunnel out of the nucleus of which they are bounded by a potential. If the correspondence principle is correct the quantum and classical probability of finding a particle in a particular position should approach each other for very high energies. 11 0 obj Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Your Ultimate AI Essay Writer & Assistant. If the particle penetrates through the entire forbidden region, it can "appear" in the allowed region x > L. The classically forbidden region is where the energy is lower than the potential energy, which means r > 2a. Here's a paper which seems to reflect what some of what the OP's TA was saying (and I think Vanadium 50 too). In particular, it has suggested reconsidering basic concepts such as the existence of a world that is, at least to some extent, independent of the observer, the possibility of getting reliable and objective knowledge about it, and the possibility of taking (under appropriate . dq represents the probability of finding a particle with coordinates q in the interval dq (assuming that q is a continuous variable, like coordinate x or momentum p). If so, how close was it? So in the end it comes down to the uncertainty principle right? We need to find the turning points where En. Remember, T is now the probability of escape per collision with a well wall, so the inverse of T must be the number of collisions needed, on average, to escape. Can you explain this answer? Acidity of alcohols and basicity of amines. Quantum mechanically, there exist states (any n > 0) for which there are locations x, where the probability of finding the particle is zero, and that these locations separate regions of high probability! probability of finding particle in classically forbidden region. Possible alternatives to quantum theory that explain the double slit experiment? probability of finding particle in classically forbidden region. We should be able to calculate the probability that the quantum mechanical harmonic oscillator is in the classically forbidden region for the lowest energy state, the state with v = 0. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. H_{2}(y)=4y^{2} -2, H_{3}(y)=8y^{2}-12y. Here you can find the meaning of What is the probability of finding the particle in classically forbidden region in ground state of simple harmonic oscillatorCorrect answer is '0.18'. Using Kolmogorov complexity to measure difficulty of problems? Description . in this case, you know the potential energy $V(x)=\displaystyle\frac{1}{2}m\omega^2x^2$ and the energy of the system is a superposition of $E_{1}$ and $E_{3}$. E is the energy state of the wavefunction. Recovering from a blunder I made while emailing a professor. The classically forbidden region is given by the radial turning points beyond which the particle does not have enough kinetic energy to be there (the kinetic energy would have to be negative). PDF | On Apr 29, 2022, B Altaie and others published Time and Quantum Clocks: a review of recent developments | Find, read and cite all the research you need on ResearchGate We turn now to the wave function in the classically forbidden region, px m E V x 2 /2 = < ()0. 2. How to match a specific column position till the end of line? This is . ample number of questions to practice What is the probability of finding the particle in classically forbidden region in ground state of simple harmonic oscillatorCorrect answer is '0.18'. Given energy , the classical oscillator vibrates with an amplitude . Bulk update symbol size units from mm to map units in rule-based symbology, Recovering from a blunder I made while emailing a professor. This expression is nothing but the Bohr-Sommerfeld quantization rule (see, e.g., Landau and Lifshitz [1981]). But for the quantum oscillator, there is always a nonzero probability of finding the point in a classically forbidden region; in other words, there is a nonzero tunneling probability. What changes would increase the penetration depth? Thus, the probability of finding a particle in the classically forbidden region for a state \psi _{n}(x) is, P_{n} =\int_{-\infty }^{-|x_{n}|}\left|\psi _{n}(x)\right| ^{2} dx+\int_{|x_{n}|}^{+\infty }\left|\psi _{n}(x)\right| ^{2}dx=2 \int_{|x_{n}|}^{+\infty }\left|\psi _{n}(x)\right| ^{2}dx, (4.297), \psi _{n}(x)=\frac{1}{\sqrt{\pi }2^{n}n!x_{0}} e^{-x^{2}/2 x^{2}_{0}} H_{n}\left(\frac{x}{x_{0} } \right) . Calculate the probability of finding a particle in the classically forbidden region of a harmonic oscillator for the states n = 0, 1, 2, 3, 4. Solution: The classically forbidden region are the values of r for which V(r) > E - it is classically forbidden because classically the kinetic energy would be negative in this case. (4.303). So it's all for a to turn to the uh to turns out to one of our beep I to the power 11 ft. That in part B we're trying to find the probability of finding the particle in the forbidden region. Such behavior is strictly forbidden in classical mechanics, according to which a particle of energy is restricted to regions of space where (Fitzpatrick 2012). Calculate the. For example, in a square well: has an experiment been able to find an electron outside the rectangular well (i.e. (ZapperZ's post that he linked to describes experiments with superconductors that show that interactions can take place within the barrier region, but they still don't actually measure the particle's position to be within the barrier region.). \[ \Psi(x) = Ae^{-\alpha X}\] 9 0 obj daniel thomas peeweetoms 0 sn phm / 0 . The calculation is done symbolically to minimize numerical errors. What is the point of Thrower's Bandolier? The classically forbidden region coresponds to the region in which. All that remains is to determine how long this proton will remain in the well until tunneling back out. 5 0 obj and as a result I know it's not in a classically forbidden region? ~! For the n = 1 state calculate the probability that the particle will be found in the classically forbidden region. Can a particle be physically observed inside a quantum barrier? Correct answer is '0.18'. To each energy level there corresponds a quantum eigenstate; the wavefunction is given by. I'm supposed to give the expression by $P(x,t)$, but not explicitly calculated. This is referred to as a forbidden region since the kinetic energy is negative, which is forbidden in classical physics. WEBVTT 00:00:00.060 --> 00:00:02.430 The following content is provided under a Creative 00:00:02.430 --> 00:00:03.800 Commons license. Classically, there is zero probability for the particle to penetrate beyond the turning points and . Are there any experiments that have actually tried to do this? b. These regions are referred to as allowed regions because the kinetic energy of the particle (KE = E U) is a real, positive value. ectrum of evenly spaced energy states(2) A potential energy function that is linear in the position coordinate(3) A ground state characterized by zero kinetic energy. << We will have more to say about this later when we discuss quantum mechanical tunneling. >> 1996-01-01. However, the probability of finding the particle in this region is not zero but rather is given by: (6.7.2) P ( x) = A 2 e 2 a X Thus, the particle can penetrate into the forbidden region. Textbook solution for Introduction To Quantum Mechanics 3rd Edition Griffiths Chapter 2.3 Problem 2.14P. Mount Prospect Lions Club Scholarship, For the particle to be found with greatest probability at the center of the well, we expect . 6 0 obj Using the change of variable y=x/x_{0}, we can rewrite P_{n} as, P_{n}=\frac{2}{\sqrt{\pi }2^{n}n! } >> There is nothing special about the point a 2 = 0 corresponding to the "no-boundary proposal". From: Encyclopedia of Condensed Matter Physics, 2005. A particle has a probability of being in a specific place at a particular time, and this probabiliy is described by the square of its wavefunction, i.e $|\psi(x, t)|^2$. Question: Probability of particle being in the classically forbidden region for the simple harmonic oscillator: a. The Franz-Keldysh effect is a measurable (observable?) "Quantum Harmonic Oscillator Tunneling into Classically Forbidden Regions", http://demonstrations.wolfram.com/QuantumHarmonicOscillatorTunnelingIntoClassicallyForbiddenRe/, Time Evolution of Squeezed Quantum States of the Harmonic Oscillator, Quantum Octahedral Fractal via Random Spin-State Jumps, Wigner Distribution Function for Harmonic Oscillator, Quantum Harmonic Oscillator Tunneling into Classically Forbidden Regions. /MediaBox [0 0 612 792] Step 2: Explanation. Is it just hard experimentally or is it physically impossible? (a) Show by direct substitution that the function, An attempt to build a physical picture of the Quantum Nature of Matter Chapter 16: Part II: Mathematical Formulation of the Quantum Theory Chapter 17: 9. Title . This is . PDF | In this article we show that the probability for an electron tunneling a rectangular potential barrier depends on its angle of incidence measured. The zero-centered form for an acceptable wave function for a forbidden region extending in the region x; SPMgt ;0 is where . Contributed by: Arkadiusz Jadczyk(January 2015) To learn more, see our tips on writing great answers. (b) Determine the probability of x finding the particle nea r L/2, by calculating the probability that the particle lies in the range 0.490 L x 0.510L . [2] B. Thaller, Visual Quantum Mechanics: Selected Topics with Computer-Generated Animations of Quantum-Mechanical Phenomena, New York: Springer, 2000 p. 168. Turning point is twice off radius be four one s state The probability that electron is it classical forward A region is probability p are greater than to wait Toby equal toe. There are numerous applications of quantum tunnelling. MathJax reference. 2. A similar analysis can be done for x 0. beyond the barrier. This is my understanding: Let's prepare a particle in an energy eigenstate with its total energy less than that of the barrier. Can you explain this answer? The number of wavelengths per unit length, zyx 1/A multiplied by 2n is called the wave number q = 2 n / k In terms of this wave number, the energy is W = A 2 q 2 / 2 m (see Figure 4-4). The turning points are thus given by En - V = 0. This is impossible as particles are quantum objects they do not have the well defined trajectories we are used to from Classical Mechanics. You'll get a detailed solution from a subject matter expert that helps you learn core concepts. /ColorSpace 3 0 R /Pattern 2 0 R /ExtGState 1 0 R But there's still the whole thing about whether or not we can measure a particle inside the barrier. /D [5 0 R /XYZ 276.376 133.737 null] Perhaps all 3 answers I got originally are the same? A measure of the penetration depth is Large means fast drop off For an electron with V-E = 4.7 eV this is only 10-10 m (size of an atom). Legal. Click to reveal HOME; EVENTS; ABOUT; CONTACT; FOR ADULTS; FOR KIDS; tonya francisco biography /Subtype/Link/A<> June 5, 2022 . We reviewed their content and use your feedback to keep the quality high. This problem has been solved! 2003-2023 Chegg Inc. All rights reserved. endobj E < V . Okay, This is the the probability off finding the electron bill B minus four upon a cube eight to the power minus four to a Q plus a Q plus. probability of finding particle in classically forbidden region \[ \tau = \bigg( \frac{15 x 10^{-15} \text{ m}}{1.0 x 10^8 \text{ m/s}}\bigg)\bigg( \frac{1}{0.97 x 10^{-3}} \]. Or am I thinking about this wrong? 2 More of the solution Just in case you want to see more, I'll . Particle always bounces back if E < V . In the ground state, we have 0(x)= m! You've requested a page on a website (ftp.thewashingtoncountylibrary.com) that is on the Cloudflare network. /Type /Annot We know that a particle can pass through a classically forbidden region because as Zz posted out on his previous answer on another thread, we can see that the particle interacts with stuff (like magnetic fluctuations inside a barrier) implying that the particle passed through the barrier. For the quantum mechanical case the probability of finding the oscillator in an interval D x is the square of the wavefunction, and that is very different for the lower energy states. [3] According to classical mechanics, the turning point, x_{tp}, of an oscillator occurs when its potential energy \frac{1}{2}k_fx^2 is equal to its total energy. The probability of the particle to be found at position x at time t is calculated to be $\left|\psi\right|^2=\psi \psi^*$ which is $\sqrt {A^2 (\cos^2+\sin^2)}$. 21 0 obj ), How to tell which packages are held back due to phased updates, Is there a solution to add special characters from software and how to do it. Published since 1866 continuously, Lehigh University course catalogs contain academic announcements, course descriptions, register of names of the instructors and administrators; information on buildings and grounds, and Lehigh history. Solution: The classically forbidden region are the values of r for which V(r) > E - it is classically forbidden because classically the kinetic energy would be negative in this ca Harmonic . Quantum tunneling through a barrier V E = T . This occurs when \(x=\frac{1}{2a}\). Such behavior is strictly forbidden in classical mechanics, according to which a particle of energy is restricted to regions of space where (Fitzpatrick 2012). endobj 30 0 obj where S (x) is the amplitude of waves at x that originated from the source S. This then is the probability amplitude of observing a particle at x given that it originated from the source S , i. by the Born interpretation Eq. A scanning tunneling microscope is used to image atoms on the surface of an object. accounting for llc member buyout; black barber shops chicago; otto ohlendorf descendants; 97 4runner brake bleeding; Freundschaft aufhoren: zu welchem Zeitpunkt sera Semantik Starke & genau so wie parece fair ist und bleibt For a classical oscillator, the energy can be any positive number. endobj . Classically, there is zero probability for the particle to penetrate beyond the turning points and . "`Z@,,Y.$U^,' N>w>j4'D$(K$`L_rhHn_\^H'#k}_GWw>?=Q1apuOW0lXiDNL!CwuY,TZNg#>1{lpUXHtFJQ9""x:]-V??e 9NoMG6^|?o.d7wab=)y8u}m\y\+V,y C ~ 4K5,,>h!b$,+e17Wi1g_mef~q/fsx=a`B4("B&oi; Gx#b>Lx'$2UDPftq8+<9`yrs W046;2P S --66 ,c0$?2 QkAe9IMdXK \W?[ 4\bI'EXl]~gr6 q 8d$ $,GJ,NX-b/WyXSm{/65'*kF{>;1i#CC=`Op l3//BC#!!Z 75t`RAH$H @ )dz/)y(CZC0Q8o($=guc|A&!Rxdb*!db)d3MV4At2J7Xf2e>Yb )2xP'gHH3iuv AkZ-:bSpyc9O1uNFj~cK\y,W-_fYU6YYyU@6M^ nu#)~B=jDB5j?P6.LW:8X!NhR)da3U^w,p%} u\ymI_7 dkHgP"v]XZ A)r:jR-4,B endobj c What is the probability of finding the particle in the classically forbidden from PHYSICS 202 at Zewail University of Science and Technology Harmonic potential energy function with sketched total energy of a particle. Once in the well, the proton will remain for a certain amount of time until it tunnels back out of the well. Estimate the tunneling probability for an 10 MeV proton incident on a potential barrier of height 20 MeV and width 5 fm. VwU|V5PbK\Y-O%!H{,5WQ_QC.UX,c72Ca#_R"n >> A particle has a certain probability of being observed inside (or outside) the classically forbidden region, and any measurements we make . % (4) A non zero probability of finding the oscillator outside the classical turning points. (a) Determine the probability of finding a particle in the classically forbidden region of a harmonic oscillator for the states n=0, 1, 2, 3, 4. 1999. The classically forbidden region is given by the radial turning points beyond which the particle does not have enough kinetic energy to be there (the kinetic energy would have to be negative). In general, quantum mechanics is relevant when the de Broglie wavelength of the principle in question (h/p) is greater than the characteristic Size of the system (d). in thermal equilibrium at (kelvin) Temperature T the average kinetic energy of a particle is . The vertical axis is also scaled so that the total probability (the area under the probability densities) equals 1. Third, the probability density distributions | n (x) | 2 | n (x) | 2 for a quantum oscillator in the ground low-energy state, 0 (x) 0 (x), is largest at the middle of the well (x = 0) (x = 0). If the proton successfully tunnels into the well, estimate the lifetime of the resulting state. In that work, the details of calculation of probability distributions of tunneling times were presented for the case of half-cycle pulse and when ionization occurs completely by tunneling (from classically forbidden region). (iv) Provide an argument to show that for the region is classically forbidden. 10 0 obj Give feedback. So which is the forbidden region. Is this possible? Summary of Quantum concepts introduced Chapter 15: 8. Peter, if a particle can be in a classically forbidden region (by your own admission) why can't we measure/detect it there? If we make a measurement of the particle's position and find it in a classically forbidden region, the measurement changes the state of the particle from what is was before the measurement and hence we cannot definitively say anything about it's total energy because it's no longer in an energy eigenstate. 1. Tunneling probabilities equal the areas under the curve beyond the classical turning points (vertical red lines). for 0 x L and zero otherwise. We have step-by-step solutions for your textbooks written by Bartleby experts! Non-zero probability to . A particle can be in the classically forbidden region only if it is allowed to have negative kinetic energy, which is impossible in classical mechanics. Classically, there is zero probability for the particle to penetrate beyond the turning points and . What is the probability of finding the particle in classically forbidden region in ground state of simple harmonic oscillatorCorrect answer is '0.18'. Wolfram Demonstrations Project Correct answer is '0.18'. endobj This is simply the width of the well (L) divided by the speed of the proton: \[ \tau = \bigg( \frac{L}{v}\bigg)\bigg(\frac{1}{T}\bigg)\] Beltway 8 Accident This Morning, Each graph depicts a graphical representation of Newtonian physics' probability distribution, in which the probability of finding a particle at a randomly chosen position is inversely related . June 23, 2022 = h 3 m k B T If the correspondence principle is correct the quantum and classical probability of finding a particle in a particular position should approach each other for very high energies. /D [5 0 R /XYZ 234.09 432.207 null] In general, we will also need a propagation factors for forbidden regions. endobj endobj How to notate a grace note at the start of a bar with lilypond? Thus, there is about a one-in-a-thousand chance that the proton will tunnel through the barrier. Non-zero probability to . xVrF+**IdC A*>=ETu zB]NwF!R-rH5h_Nn?\3NRJiHInnEO ierr:/~a==__wn~vr434a]H(VJ17eanXet*"KHWc+0X{}Q@LEjLBJ,DzvGg/FTc|nkec"t)' XJ:N}Nj[L$UNb c /Subtype/Link/A<> $x$-representation of half (truncated) harmonic oscillator? 06*T Y+i-a3"4 c 162.158.189.112 << "After the incident", I started to be more careful not to trip over things. Qfe lG+,@#SSRt!(` 9[bk&TczF4^//;SF1-R;U^SN42gYowo>urUe\?_LiQ]nZh This Demonstration shows coordinate-space probability distributions for quantized energy states of the harmonic oscillator, scaled such that the classical turning points are always at . This should be enough to allow you to sketch the forbidden region, we call it $\Omega$, and with $\displaystyle\int_{\Omega} dx \psi^{*}(x,t)\psi(x,t) $ probability you're asked for. Can you explain this answer? Can you explain this answer? To find the probability amplitude for the particle to be found in the up state, we take the inner product for the up state and the down state. We have so far treated with the propagation factor across a classically allowed region, finding that whether the particle is moving to the left or the right, this factor is given by where a is the length of the region and k is the constant wave vector across the region.