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The TISE is given by $$i\hbar\frac{d}{dt}|\psi(t)\rangle=\hat{H}|\psi(t)\rangle$$ Where $|\psi(t)\rangle\in\mathcal{H}$ is a state in the Hilbert space $\mathcal{H}$. Now, in the position basis, the TISE is given by (WLOG. consider the 1 dimensonal case). $$i\hbar\frac{\partial}{\partial t}\Psi(x,t)=\hat{H}\Psi(x,t)$$ Where $$\Psi(x,t)=\langle x|\psi(t)\rangle.$$ I assume that because the first equation is more fundamental, it should be possible to derive the second from it. I thus took the inner product of the first equation with the bra $\langle x|$. The LHS rather easily reduces to the second equation by the very definition of $\Psi(x,t)$, but I had trouble with the RHS and the Hamiltonian operator. I know that $$\langle x|\hat{H}\psi(t)\rangle=\langle \hat{H}x|\psi(t)\rangle,$$ but couldn't advance further.

How would you show that $$\langle x|\hat{H}\psi(t)\rangle=\hat{H}\langle x|\psi(t)\rangle=\hat{H}\Psi(x,t)~?$$ Is this even a sensible approach?

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Hamiltonian in both cases acts on different objects so it must be different mathematical entity. The trick is $$ \left<x\right|\hat{H}\left|\Psi(t)\right>=\int dx'\left<x\right|\hat{H}\left|x'\right>\left<x'\right|\left.\Psi(t)\right>=\left<x\right|\hat{H}\left|x\right>\left<x\right|\left.\Psi(t)\right> $$

Edit: The second equality Holds if hamiltonian is diagonal in position basis. That this is true for standard form hamiltonian can be shown computing matrix elements:

$$ H_{xx'}=\left<x\right|\hat{H}\left|x'\right>=\left<x\right|\left(\frac{\hat{P}^2}{2m}+\hat{V}\right)\left|x'\right>=\left<x\right|\frac{\hat{P}^2}{2m}\left|x'\right>+\left<x\right|\hat{V}\left|x'\right> $$ The potential term is diagonal if it is defined as function of position operator $ \hat{x}\left|x\right>=x\left|x\right>, $ which it usually is and this operator is by definition diagonal in position basis.

Less trivial is the momentum operator, which is defined as generator of translation: $$\left|x+dx\right>=\left(1-i\hat{P}dx/\hbar\right)\left|x\right>$$ From this definition we can compute: $$ \left<x\right|\hat{P}\left|x'\right>=i\hbar\frac{\left<x\right.\left|x'\right>-\left<x\right.\left|x'-dx''\right>}{dx''} $$ In the limit $dx''\rightarrow 0$ this shows, that momentum operator is indeed diagonal in position basis, albeit there is distribution on the diagonal instead of ordinary numbers as it was in the case of potential. And since momentum operator is diagonal, so it is its second power and thus also the kinetic term of hamiltonian.

To continue the journey to the standard equation for wave function we can write for momentum $$ \int dx'\left<x\right|\hat{P}\left|x'\right>\left<x'\right|\left.\Psi(t)\right>=i\hbar\int dx'\frac{\left<x\right.\left|x'\right>-\left<x\right.\left|x'-dx''\right>}{dx''}\left<x'\right|\left.\Psi(t)\right>=i\hbar\int dx'\frac{\delta(x-x')-\delta(x-x'+dx'')}{dx''}\left<x'\right|\left.\Psi(t)\right>=i\hbar\frac{\Psi(x,t)-\Psi(x+dx'',t)}{dx''} \rightarrow -i\hbar\frac{d}{dx}\Psi(x,t) $$ In similar fashion you can get the second power of momentum operator and then adding the potential function you get standard hamiltonian in wave mechanics.

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  • $\begingroup$ Are you assuming $\hat{H}$ is diagonal in the position basis or have you just lost some primes in the last equation? $\endgroup$ – jacob1729 Jun 3 at 10:37
  • $\begingroup$ yes i do assume diagonality $\endgroup$ – Umaxo Jun 3 at 11:57
  • $\begingroup$ I'm guessing that you used the projection operator $\mathbb{1}=\int dx' |x'\rangle\langle x'|$ to get the first equality, but I'm not sure how you got the second one? $\endgroup$ – Zachary Jun 3 at 17:08
  • $\begingroup$ @Zachary i expanded the answer in the edit $\endgroup$ – Umaxo Jun 3 at 18:53
  • $\begingroup$ I really don't understand this answer. At the end of the day you can just say $H=T+V$ and calculate their scalar product with $\langle x|$ independently. All the integrals that are shown are not needed. $\endgroup$ – gented Jun 3 at 19:21
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$$\left<x\right|\hat{H}\left|\Psi(t)\right>=\left<x\right|\left(\frac{\hat{P}^2}{2m}+{V(\hat x)}\right)\left|\Psi(t)\right>=\left<x\right|\frac{\hat{P}^2}{2m}\left|\Psi(t)\right>+\left<x\right|{V(\hat x)}\left|\Psi(t)\right>$$ Let's evaluate $\left<x\right|\frac{\hat{P}^2}{2m}\left|\Psi(t)\right>$ by completeness property. $$\left<x\right|\frac{\hat{P}^2}{2m}\left|\Psi(t)\right>=\left<x\right|\frac{\hat{P}.\hat{P}}{2m}\left|\Psi(t)\right>={1\over 2m}\int dx'\left<x\right|\hat{P}\left|x'\right>\left<x'\right|\hat{P}\left|\Psi(t)\right>$$ Use the results

$$\left<x\right|\hat{P}\left|x'\right>=i\hbar {\partial \over \partial{x'}}\delta(x'-x)$$ $$\left<x'\right|\hat{P}\left|\Psi(t)\right>=-i\hbar {\partial \over \partial{x'}}\left<x'|\Psi(t)\right>$$ If you are interested in the proof of these things, I will prove it later.

$$\left<x\right|\frac{\hat{P}^2}{2m}\left|\Psi(t)\right>={1\over 2m}\int dx'i\hbar {\partial \over \partial{x'}}\delta(x'-x)i\hbar {\partial \over \partial{x'}}\left<x'|\Psi(t)\right>={\hbar ^2 \over 2m}\int dx' {\partial \over \partial{x'}}\left<x'|\Psi(t)\right>{\partial \over \partial{x'}}\delta(x'-x)$$ Applying product rule for differentiation $${\hbar ^2 \over 2m}\int dx' {\partial \over \partial{x'}}\left<x'|\Psi(t)\right>{\partial \over \partial{x'}}\delta(x'-x)={-\hbar ^2 \over 2m}\int dx' \delta(x'-x) {\partial \over \partial{x'}}{\partial \over \partial{x'}}\left<x'|\Psi(t)\right>$$ We get $$\left<x\right|\frac{\hat{P}^2}{2m}\left|\Psi(t)\right>={-\hbar ^2 \over 2m}{\partial ^2 \over \partial{x^2}}\left<x|\Psi(t)\right>$$ Similarly $$\left<x\right|{V(\hat x)}\left|\Psi(t)\right>=V(x)\left<x|\Psi(t)\right>$$ Combining these together, $$\left<x\right|\hat{H}\left|\Psi(t)\right>=\left<x\right|\frac{\hat{P}^2}{2m}\left|\Psi(t)\right>+\left<x\right|{V(\hat x)}\left|\Psi(t)\right>={-\hbar ^2 \over 2m}{\partial ^2 \over \partial{x^2}}\left<x|\Psi(t)\right>+V(x)\left<x|\Psi(t)\right>=\Biggr ({-\hbar ^2 \over 2m}{\partial ^2 \over \partial{x^2}}+V(x)\Biggl )\left<x|\Psi(t)\right>=\hat{H}\left<x|\Psi(t)\right>$$ $$\left<x\right|\hat{H}\left|\Psi(t)\right>=\hat{H}\left<x|\Psi(t)\right>$$ Proofs:

$$\left<x\right|\hat{P}\left|x'\right>=i\hbar {\partial \over \partial{x'}}\delta(x'-x)$$

Use the commutation relation $$\left<x\right|[\hat{x},\hat{P}]\left|x'\right>=i\hbar {\left<x|x'\right>} $$ $$\left<x\right|[\hat{x},\hat{P}]\left|x'\right>=\left<x\right|\hat {x}\hat{P}-\hat {P}\hat {x} \left|x'\right>=(x-x')\left<x\right|\hat{P}\left|x'\right>\implies$$ $${i\hbar {\left<x|x'\right>}\over (x-x')}=\left<x\right|\hat{P}\left|x'\right> $$ $$\left<x\right|\hat{P}\left|x'\right>={i\hbar \delta(x'-x) \over (x-x') }=i\hbar {\partial \over \partial{x'}}\delta(x'-x)$$

$$\left<x'\right|\hat{P}\left|\Psi(t)\right>=-i\hbar {\partial \over \partial{x'}}\left<x'|\Psi(t)\right>$$

Let $$\left<x'\right|\hat{P}\left|\Psi(t)\right>=\int dx \left<x'\right|\hat{P}\left|x\right> \left<x|\Psi(t)\right>$$ Apply the previous result $$\left<x'\right|\hat{P}\left|\Psi(t)\right>=i\hbar\int dx \left<x|\Psi(t)\right>{\partial \over \partial{x}}\delta(x-x')$$ Use Product rule for differentiation as we done before, Hence obtain $$\left<x'\right|\hat{P}\left|\Psi(t)\right>=-i\hbar {\partial \over \partial{x'}}\left<x'|\Psi(t)\right>$$ Edit:

$$\delta '(x-a)={-\delta (x-a)\over x-a}$$

See the pdf below.

https://www.google.com/url?sa=t&source=web&rct=j&url=http://physicspages.com/pdf/Mathematics/Derivatives%2520of%2520delta%2520function.pdf&ved=2ahUKEwinwoizp9biAhUFmuYKHTgQASgQFjAIegQIARAB&usg=AOvVaw3CsWpXDPpm1EIZcKBpGRTN

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    $\begingroup$ What a beautiful calculation for a (seemingly) simple equality!! $\endgroup$ – descheleschilder Jun 4 at 8:45
  • $\begingroup$ Thank you for the detailed answer. Could you explain why $$\frac{\delta(x'-x)}{x-x'}=\frac{\partial}{\partial x'}\delta(x'-x)$$ $\endgroup$ – Zachary Jun 5 at 13:32
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    $\begingroup$ "Delta function is continuous at all its point. So it's derivative exist." That's kind of problematic. Not only that the delta function is not continuous (let alone a function), even in general continuity does not imply differentiability. $\endgroup$ – eranreches Jun 6 at 6:05
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    $\begingroup$ @walber97 It is not so trivial. I would just state it as a property, as one does with all the other formulas of the delta function. A motivation as to why it is true is done only under the integral sign. The 'proofs' you are given as a physicist (including this one) are not really valid. $\endgroup$ – eranreches Jun 6 at 19:06
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    $\begingroup$ @walber97 BTW, this Wikipedia page might be useful en.wikipedia.org/wiki/…. $\endgroup$ – eranreches Jun 7 at 2:43

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