Schrödinger Wave Equation: Tashin da Amfani da Karamin Tsarin Littattafai

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Wani Schrodinger Equation?

Tana Schrodinger equation (ko kuma ake nufin Schrödinger’s wave equation) wani matsayin tushen kafin kafin kafuwar jikin kwantum ta hanyar tushen waya. Tarihin, abubuwan da ke tsakiya, da kuma energy cikin wannan jiki zai iya samun da shi a kan bayyana Schrödinger equation.

Duk fahimtarsa ta zuba masu jiki na kafuwar kwantum suna cikin tushen waya. Tushen waya zai ba da nasara a kan bayyana da kuma samun Schrodinger equation. Schrodinger equation wani babban aksiomi da ake bar da ita a cikin littattafan physics na biyu. Yana da kyau a gano a kan littattafan electrical engineering a jami'o'i saboda yana iya amfani da ita a semiconductors.

Amma, a cikin duka waɗannan, ana nuna ita ne a kan bayyana kawai, kuma ba ake gina a kan halin da ya fi kyau. Wannan yana da karfi saboda duk wadannan da ake karatu a cikin physics na biyu ta kwantum suka gudanar da wannan asasinsa. A cikin takardun, za a gina ita daga baya da zan yi aiki da ake nuna duk ƙarin da aka yi.

Zan iya cewa, ƙarin da zan yi suna daidai da ƙarin da Schrödinger ya yi, saboda haka za ku iya duba ƙarin da mai sarauta ya yi a zamansa. A nan, haka ne Schrödinger equation na lokacin da yawa a cikin 3-dimensions (wajen mutanen da ba su kaɗansu) a cikin haɗin:

Physics na Kwantum da Waya

Duka sun ce da suke ƙare physics na biyu – amma ya jagoranci minkali (koyar da Newtonian mechanics, Maxwell’s equations, da kuma special relativity).

Amma, kamar yadda ake nuna a cikin labarai da suka rasa, abubuwan da aka samu a tarihin layi ba su zama na tsawon hankali idan ta haka mutane sun sanin fahimta masu ilimi a lokacin. Labarai masu tsohon magana da kuma to some degree photoelectric effect sun fi siffar da abubuwa masu gaba-gaban da ba su zama na tsawo da fahimta masu ilimi a lokacin.

Amma, yaushe? A matsayin bayani, a ilimin lissafi ana iya tabbatar da duwatsuwa, jiki da magana. Muhimmanci da na iya bayyana game da wannan duwatsuwa masu gaba-gaban shine kamar yadda ake nuna:

Jiki: duka mai karfi da shi ne da karami da kula $m$ .

Magana: yawan alamomin da ke faruwa a cikin wani wurin da ke faruwa a lokacin. Ana iya bayyana su da funfaruwar magana $\psi(\vec{r}, t)$ wanda ya bayyana funfaruwar magana a cikin wurin da lokacin.

Wannan ya bazu mu zuwa abubuwan da suka samun daidai a cikin Photoelectric Emission. An samu cewa electron ya nuna duwatsuwa masu gaba-gaban. Wannan ya haɗa da fahimtar da suka sanin lokacin, saboda an yi nasara da wannan duwatsuwa masu gaba-gaban da suka zama na hankali.

Babu na musamman? A lokacin, wasu mutane da suka fiye a cikin ilimin lissafi suka samu cewa akwai kungiyoyi a cikin ilimi, da kuma babban nasara ta faru a lokacin Louis de Broglie ya kula kan momentum (da ke jiki) zuwa wavelength (da ke magana) da ke bayyana da

$\begin{equation*} p = h/\lambda. \end{equation*}$

Kuma, daga Photoelectric Emission a nan da shi cewa abin hawa da fitarwa na photons (wannan yana da shakka cewa shine zarra ko kuma hawa) suna da abin da ya ba da

$\begin{equation*} E = hf = \hbar \omega \end{equation*}$

Idan $\hbar = h/2\pi$ da $\omega=2\pi f$ . Tana da muke so kuɗi da Schrödinger ke ciki saboda takaice ta. Amma inda za su faru? Ba gaskiya ne cewa electrons da photons sun nuna nasara a tsawo da nasara a yanayi. Bana da wani abu da zai iya jawabi waɗanda duka tsawon mutane suke da suke don samun ƙarin bayani.

Yadda a Haɗa Tsawon Kima

Girman $\psi(\vec{r}, t)$ yana da tsawon kima. Zanen, electron yana nuna nasara a tsawo da kuma nasara a yanayi. Saboda haka, a lokacin da zuwa, ina bincike maɗanin electromagnetic fields. A wannan halin, Maxwell’s equations sun haɗa da su da suka da su.

$\begin{align*} \nabla \times \vec{E} &= - \frac{\partial{\vec{B}}}{\partial{t} } \\ \nabla \times \vec{B} &= -\mu_0 \left(\vec{J} + \epsilon_0\frac{\partial{\vec{E}}}{\partial{t}} \right)\\ \nabla \cdot \vec{E} &= \frac{\rho}{\epsilon_0}\\ \nabla \cdot \vec{B} &= 0 \end{align*}$

Idan $c$ ya shi da sauyin cikin kungiyar, $\vec{E}$ ya shi da sauti na farafin, da $\vec{B}$ ya shi da sauti na magana. Tushen da aka bayar a yanzu shi babban tushen da ake amfani a matsayin gine-gine na farafin, indakatauna, da kuma transformers, kuma shi ne mai karatuwa na Faraday.

Kuma, wanda daga cewa $\nabla \cdot \vec{B} = 0$ ba da sauki da za suka ci gaba ita ce bai ci da monopole na magana. Fahimtar hanyar samun waɗannan tushen da ma'anarsu a kanadanci yana ba da inganci aiki a kan injinirin. A nan, zan iya samun tushen da duk alamar electromagnetism suna bi da shi a fada:

$\begin{align*} \nabla \times \vec{E} &= - \frac{\partial{\vec{B}}}{\partial{t} }\\ \implies \nabla \times (\nabla \times \vec{E}) &= - \frac{\partial{(\nabla \times \vec{B})}} {\partial{t} }\\ \implies \nabla \times (\nabla \times \vec{E}) &= -\frac{1}{c^2} \frac{\partial^2{\vec{E}}}{\partial{t^2} } \end{align*}$

A nan za a iya amfani da idan da ake sanin da kuma ake daidaita (da ake iya tabbatar da shi): $\nabla \times (\nabla \times T) = \nabla(\nabla \cdot T) - \nabla^2T$ idan $T$ ya shi da saukan da ake amfani a kan. Amfani da wannan a kan tushen da ake bayar a nan:

$\begin{align*} \nabla(\nabla \cdot \vec{E}) - \nabla^2 \vec{E} &= -\frac{1}{c^2} \frac{\partial^2{\vec{E}}}{\partial{t^2} }\\ \implies - \nabla^2 \vec{E} &= -\frac{1}{c^2}\frac{\partial^2{\vec{E}}}{\partial{t^2} }\\ \nabla^2 \vec{E} - \frac{1}{c^2}\frac{\partial^2{\vec{E}}}{\partial{t^2}} & = 0 \end{align*}$

Na bi da ya fi ake koyar da wannan maida a nan, shi ne maida mai tsawon lashe-rashe a tsohon harkokin diminsin uku. Wannan maida yana nuna a cikin lashe-rashe na farko, amma yana iya zama a cikin lashe-rashe na akwai, lashe-rashe na zamani, lashe-rashe na takaitaccen ruwa, da kuma lashe-rashe na al'adu.

Daidaita Maida Schrödinger

Bayanan Lashe-Rashe Na Tsohon Lasa

Tun daga maida lashe-rashe na tsohon lasa (yana da kyau a yi aiki a tsohon diminsi baya bayan hakan don in iya faɗa zuwa tsohon diminsi uku saboda logikan yana iya haɗa a duk: $x, y$ , da $z$ ):

$\begin{equation*} \frac{ {\partial^2{E}} }{\partial^2{x}} = \frac{1}{c^2} \frac{ {\partial^2{E}} }{\partial^2{t}} \Longrightarrow \frac{ {\partial^2{E}} }{\partial^2{x}} - \frac{1}{c^2} \frac{ {\partial^2{E}} }{\partial^2{t}} = 0 \end{equation*}$

Wannan ita ce, a fanni, maida mai tsawon kammalawa na wuce da yake amsa a matsayin bayanan lashe-rashe na tsohon lasa:

$\begin{equation*} E(x, t) = E_0 e^{i(kx - \omega t)} \text{ (check this for yourself!). } \end{equation*}$

Daga kuma na sani da wasu kungiyoyin hanyar tsakiyar mafi yawa cewa $k= \frac{2\pi}{\lambda}$ da $\omega = 2 \pi f$ . Yanzu, za mu iya amfani da abubuwa daga Einstein da Compton kuma zan iya rubuta cewa kyauwar fotoni shi ne $\mathsf{E} = \hbar \omega$ da kuma daga de-Broglie cewa $p = h / \lambda = \hbar k$ . Zan iya taka wani karin bincike a fili na jihohi:

$\begin{equation*} E(x, t) = E_0 e^{\frac{i}{\hbar}(px - \mathsf{E} t)} \end{equation*}$

Wannan shi ne kyakkyawan jihohi mai bayyana fotoni. Za mu iya rubuta wannan kyakkyawan a cikin kyakkyawan tsakiyar mafi yawa da gaskiya muna neman!

$\begin{align*} \left(\frac{ {\partial^2{}} }{\partial^2{x}} - \frac{1}{c^2} \frac{ {\partial^2{}} }{\partial^2{t}}\right) E_0 e^{\frac{i}{\hbar}(px - \mathsf{E} t)} &= 0\\ \implies -\frac{1}{\hbar^2} \left( p^2 - \frac{\mathsf{E} ^2}{c^2} \right) E_0 e^{\frac{i}{\hbar}(px - \mathsf{E} t)} &= 0 \end{align*}$

A cikin haka, $\mathsf{E}^2 = p^2 c^2$ wanda ya fi kyau saboda ana sani daga ilimin tattalin kasa cewa energy na yawa da kuma massa $m$ shine:

$\begin{equation*} \mathsf{E}^2 = p^2c^2 + m^2 c^4 \end{equation*}$

Kuma muna yi aiki da photon ne kawai, wanda ba ta da massa $(m=0)$ ! Saboda haka, za a iya samun fahimtata da za a iya amfani da energy na yawa da kuma massa (kamar electron misali) da kuma zama sunan equation na zuwa $\Psi$ saboda muna yi aiki.

$\begin{equation*} -\frac{1}{\hbar^2} \left( p^2 - \frac{\mathsf{E}^2}{c^2} + m^2c^2 \right) \Psi e^{\frac{i}{\hbar}(px - \mathsf{E} t)} = 0 \end{equation*}$

Wannan equation ya faru saboda amfani da plane wave equation da ke fito da photon a wave equation. Amma, saboda muna son in bincika energy na yawa da kuma massa, muna bukatar da gwadon wave equation. Wannan shine saboda wave equation ba zai iya haɗa da $\Psi$ wanda ake amfani da shi don bayyana particles da waves. Muna iya samun operator don samun equation na zuwa, da ke gaba:

$\begin{equation*} \left( \frac{ {\partial^2{}} }{\partial^2{x}} - \frac{1}{c^2} \frac{ {\partial^2{}} }{\partial^2{t}} - \frac{m^2c^2}{\hbar^2} \right)\Psi e^{\frac{i}{\hbar}(px - \mathsf{E} t)} = 0 \end{equation*}$

Yadda Amsa Masu Yawan Kafi Da Turanci Da Tsarin Turanci

Na gane da kuma yin zama da ake bayyana da shi a cikin $\mathsf{E}$ masu harufa da yawan kafi. Za a yi karfin yin tsari don a iya amfani da wasu zama.

$\begin{align*} \mathsf{E} ^2 &= p^2c^2 + m^2c^4\\ \mathsf{E} &= \sqrt{\left( p^2c^2 + m^2c^4 \right)}\\ &= \sqrt{\left( c^4(\frac{p^2}{c^2} + m^2) \right)}\\ &= \sqrt{\left( c^4 m^2(\frac{p^2}{m^2 c^2} + 1) \right)}\\ &= mc^2\sqrt{\left(\frac{p^2}{m^2 c^2} + 1 \right)} \end{align*}$

Gaskiya na biyu wannan babban zama shine ya fara da tsari zuwa hakan $\sqrt{1 + x}$ saboda idan a yi karfin Taylor Series na wannan tsari za a samun:

$\begin{equation*} \sqrt{1 + x} \approx 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} + ... \end{equation*}$

Idan $x$ yana da shi, wani babban kofin da ke cikin taylors expansion shine $O(1)$ . A cikin formular masu energy na, $x = \frac{p^2}{m^2 c^2 } =\left( \frac{p}{mc }\right)^2$ . Zan iya amfani da abubuwa na $p = mv \ll mc$ saboda haka, wanda ba a yi ziyarar da light speed (na zo in tabbataki idan ka samu wani abu da ba sa!)! Don haka wannan term ya zama:

$\begin{align*} \mathsf{E} &= mc^2\sqrt{\left(\frac{p^2}{m^2 c^2} + 1 \right)}\\ & \approx mc^2 \left( 1 + \frac{1}{2} \frac{p^2}{m^2 c^2} \right)\\ & = mc^2 + \frac{p^2}{2m} = mc^2 + E_{\text{kinetic}} \end{align*}$

Daga

$\begin{equation*} E_\text{kinetic} = \frac{1}{2} mv^2 = \frac{1}{2} \frac{(mv)^2}{m} = \frac{p^2}{2m} \end{equation*}$

Shi ne kinetic energy na biyu da ake gano a high school physics. Hadi karin wave function daga zuwa, zan iya amfani da wannan bayanin da aka bayar don in tafi:

$\begin{align*} \Psi(\vec{r},t) &= \Psi_0 e^{\frac{i}{\hbar}(p \vec{r} - \mathsf{E} t)}\\ &= \Psi_0 e^{\frac{i}{\hbar}(p\vec{r} - mc^2t - E_{\text{kinetic}}t)}\\ &= e^{-\frac{i}{\hbar}mc^2t} \Psi_0 e^{\frac{i}{\hbar}(p\vec{r} - E_{\text{kinetic}}t)}\\ \end{align*}$

Sabbin da yadda muka kawo sunan abubuwa biyu a kan ina ya shafi cewa babban sunan $e^{-\frac{i}{\hbar}mc^2t}$ (wanda ake kira saboda hirin na tsaye) zai yi wakarri a gaba da fadin sunan biyu kuma ba zai bayyana abubuwan da muke neman baya. Saboda haka, don in taimaka da wannan farkon, za mu ci:

$\begin{equation*} \Psi(\vec{r},t) = e^{-\frac{i}{\hbar}mc^2t} \psi(\vec{r}, t) \end{equation*}$

Daga nan, muna bayyana:

$\begin{equation*} \psi(\vec{r}, t) =\Psi_0 e^{\frac{i}{\hbar}(p\vec{r} - E_{\text{kinetic}}t)}. \end{equation*}$

Amsa, za mu iya samun birnin da takamin birnin $\Psi(\vec{r},t)$ kuma za mu iya gani wadannan. Amsa:

$\begin{equation*} \frac{\partial{\Psi}}{\partial t} = -\frac{i}{\hbar}mc^2e^{-\frac{i}{\hbar}mc^2t} \psi(\vec{r}, t) + e^{-\frac{i}{\hbar}mc^2t} \frac{\partial \psi(\vec{r}, t)}{\partial t} \end{equation*}$

da ta mi:

$\begin{equation*} \frac{\partial^2{\Psi}}{\partial t^2} = \left( -\frac{m^2c^4}{\hbar^2} e^{-\frac{i}{\hbar}mc^2t}\psi - \frac{2i}{\hbar}mc^2e^{-\frac{i}{\hbar}mc^2t}\frac{\partial \psi}{\partial t} \right) + e^{-\frac{i}{\hbar}mc^2t}\frac{\partial^2 \psi}{\partial t^2} \end{equation*}$

Yana da kyau a lura cewa abubuwan da na biyu a matsayin hanyar tattalin kammal mai yawa saboda ba a bani $c^2$ wanda ke da muhimmanci, kuma don haka a kan bayanai, tattalin kammal mai yawa shi ne:

$\begin{align*} \frac{\partial^2{\Psi}}{\partial t^2} \approx \left( -\frac{m^2c^4}{\hbar^2} e^{-\frac{i}{\hbar}mc^2t}\psi - \frac{2i}{\hbar}mc^2e^{-\frac{i}{\hbar}mc^2t}\frac{\partial \psi}{\partial t} \right) \end{align*}$

Sababinsu da muna gano waɗannan abubuwa biyu a matsayin hanyar tattalin kammal shine don in yi amfani da su a wannan batun da ke nuna tattalin kammal mai yawa:

Amma a nan, zan iya yi haka, zan iya dogara wannan batun, kuma zan samu batun da ake kira batun ta Klein-Gordon:

$\begin{align*} \left( \frac{ {\partial^2{}} }{\partial^2{x}} - \frac{1}{c^2} \frac{ {\partial^2{}} }{\partial^2{t}} - \frac{m^2c^2}{\hbar^2} \right)\Psi_0 e^{\frac{i}{\hbar}(px - \mathsf{E} t)} &= 0\\ \frac{ {\partial^2{\Psi(x, t)}} }{\partial^2{x}} - \frac{m^2c^2}{\hbar^2} \Psi(x, t) &= \frac{1}{c^2} \frac{ {\partial^2{\Psi(x, t)}} }{\partial^2{t}} \end{align*}$

A nan da zaka iya kawo wannan zuwa 3-dimensions tare da koyar da wannan ta hanyar bayanin vector (dukkan cikakken da ake amfani da su don samun wannan formular za su iya amfani da su a kan duka $x,y$ , da $z$ .)

$\begin{equation*} \nabla^2 \Psi(\vec{r}, t) - \frac{m^2c^2}{\hbar^2} \Psi(\vec{r}, t) = \frac{1}{c^2} \frac{ {\partial^2{\Psi(\vec{r}, t)}} }{\partial^2{t}} \end{equation*}$

Wannan bayanin ya shahara a cikin sunan bayanin Klein-Gordon wanda yake ke free particle. Wannan bayani na relativistic saboda babban abubuwan da ke aiki ba ta yi amincewa da cewa ake yi a kan little $\sqrt{1+x}$ Taylor expansion.

Yanzu, bari in koyar da bayanin Klein-Gordon (za a gaba zuwa 1-D da koyar da wannan energy formula) da muna samu Schrödinger Equation:

$\begin{align*} \frac{ {\partial^2{\Psi}} }{\partial^2{x}} - \frac{m^2c^2}{\hbar^2} \Psi &= \frac{1}{c^2} \frac{ {\partial^2{\Psi}} }{\partial^2{t}} \end{align*}$

Bari in fadada wannan wave function da ke bayar da $\Psi(\vec{r},t) = e^{-\frac{i}{\hbar}mc^2t} \psi(\vec{r}, t)$ inda muna sani cewa me yadda first and second derivatives with respect to time yana nufin:

$\begin{align*} \frac{ {\partial^2{}} }{\partial^2{x}}e^{-\frac{i}{\hbar}mc^2t} \psi - \frac{m^2c^2}{\hbar^2} e^{-\frac{i}{\hbar}mc^2t} \psi &= \frac{1}{c^2}\left( -\frac{m^2c^4}{\hbar^2} e^{-\frac{i}{\hbar}mc^2t}\psi - \frac{2i}{\hbar}mc^2e^{-\frac{i}{\hbar}mc^2t}\frac{\partial \psi}{\partial t} \right) + e^{-\frac{i}{\hbar}mc^2t}\frac{\partial \psi}{\partial t}\\ \frac{ {\partial^2{}} }{\partial^2{x}}e^{-\frac{i}{\hbar}mc^2t} \psi &= \frac{m^2c^2}{\hbar^2} e^{-\frac{i}{\hbar}mc^2t} \psi -\frac{m^2c^2}{\hbar^2} e^{-\frac{i}{\hbar}mc^2t}\psi - \frac{2i}{\hbar}me^{-\frac{i}{\hbar}mc^2t}\frac{\partial \psi}{\partial t} + e^{-\frac{i} {\hbar}mc^2t}\frac{\partial^2 \psi}{\partial t^2}\\ \frac{ {\partial^2{}} }{\partial^2{x}}e^{-\frac{i}{\hbar}mc^2t} \psi &= -\frac{2i}{\hbar}me^{-\frac{i}{\hbar}mc^2t}\frac{\partial \psi}{\partial t} \\ e^{-\frac{i}{\hbar}mc^2t}\left( \frac{ {\partial^2{\psi}} }{\partial^2{x}} +\frac{2im}{\hbar}\frac{\partial \psi}{\partial t} \right) &= 0 \end{align*}$

Yanzu idan da yadda ake yi shine koyarwa ta musamman don samun Tashar Schrödinger a uku (karamin hada cewa $\frac{1}{i} = -i$ ):

$\begin{equation*} i \hbar \frac{\partial{}}{\partial{t}} \Psi(\vec{r},t) = \frac{-\hbar^2}{2 m} \nabla^2 \Psi(\vec{r},t) \end{equation*}$

Amsa ake iya bayyana da nuna hakan cewa alama a kan babban tushen tashar yana nufin jumlar masu sauti.

A cikin bayyana na, ake sani cewa $V(\vec{r},t)$ ya shiga 0 kuma ake yi amfani da karamin zafi kawai. A nan ake sani cewa alama ta kiyaye daidai a kan takaitaccen abubuwa, saboda haka, Tashar Schrödinger a uku da alama ta shi shine:

$\begin{equation*} i \hbar \frac{\partial{}}{\partial{t}} \Psi(\vec{r},t) = \left[\frac{-\hbar^2}{2 m} \nabla^2 +V(\vec{r},t)\right] \Psi(\vec{r},t). \end{equation*}$

Na'am! Duk da za a iya bayyana Tashar Schrodinger na mutanen da ba su dace ba a uku. Idan kana son samun bayanan da ke so, zaka iya karatu muna biyuwar da za a iya bayyana bayanan da ke so.

Nadamanin

Gasiorowicz, S. (2019). Fisika Kwantum. 2nd ed. Canada: Hamilton Printing, pp.1-50.
Griffiths, D. (2019). Fisika Kwantum. 3rd ed. University Printing House, Cambridge: Cambridge University Press.
Ward, D. and Volkmer, S. (2019). Yadda A Gyara Tsarin Schrodinger. [online] arXiv.org. Available at: https://arxiv.org/abs/physics/0610121v1 [Accessed 29 May 2019].
Shankar, R. (1980).Principles of Quantum Mechanics. 1st ed. New York: Springer Science, pp.1-40.

Bayanin: Gaskiya wani, abubuwa mai kyau za su iya shirya, idana da wani hankali don bayyana, zaka so in bace.

Ba da kyau kuma kara mai rubutu!

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