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I want to check that I am getting the concept right here, and my question is: if the expectation value of a Hamiltonian is the same whether you use the time dependent version or not. I thought I had it right initially -- maybe I did -- but I wanted to make sure I didn't go off the rails somewhere.

We have a wave function: $\psi = \alpha \phi_1 + \beta \phi_2$ and normalized it's $\frac{1}{(\alpha^2 + \beta^2)}(\alpha \phi_1 + \beta \phi_2)$.

The time evolution of that state will be $(\frac{1}{\alpha^2 + \beta^2})(\alpha \phi_1 e^{-i\omega_1 t} + \beta \phi_2 e^{-i\omega_2 t})$

And $\hat H \psi = i\hbar \frac{\partial}{\partial t}\psi(x,t)$ and $\langle \hat H \rangle = \int \psi \hat H \psi dx$

$\frac{\partial \psi}{\partial t} = \frac{1}{\alpha^2 + \beta^2} (-i\omega_1 \alpha \phi_1 e^{-i\omega_1 t}-i\omega_2 \beta e^{-i\omega_2 t}\phi_2)$

So the integral is $$\int \psi \hat H \psi dx = \int (\frac{1}{\alpha^2 + \beta^2})(\alpha \phi_1 e^{-i\omega_1 t} + \beta \phi_2 e^{-i\omega_2 t})i \hbar (-i\omega \alpha \phi_1 e^{-i\omega_1 t}-i\omega_2 \beta e^{-i\omega_2 t}\phi_2)dx$$ $$=-\hbar\frac{1}{\alpha^2 + \beta^2}\int(\alpha \phi_1 e^{-i\omega_1 t} + \beta \phi_2 e^{-i\omega_2 t}) (\alpha \omega_1 \phi_1 e^{-i\omega_1 t}-\beta \omega_2e^{-i\omega_2 t}\phi_2)dx$$ $$=-\hbar\frac{1}{\alpha^2 + \beta^2}\int (\alpha^2 \omega_1 \phi_1^2 e^{-2i\omega_1 t}-\beta^2 \omega_2 \phi_2^2e^{-2i\omega_2 t}\phi_2)dx$$

Using a 0 to L limit (we're doing a particle in a box here), and taking the sinusoidal form of $\phi_n$: $$=\frac{-2\hbar}{L (\alpha^2 + \beta^2)}\int^L_0 \alpha^2 \omega_1 e^{-2i\omega_1 t}\sin^2({\frac{\pi x}{L})}-\beta^2 \omega_2 e^{-2i\omega_2 t}\sin^2({\frac{2\pi x}{L})}dx$$

applying a trig identity

$$=\frac{-2\hbar}{L (\alpha^2 + \beta^2)}\int^L_0 \alpha^2 \omega_1 e^{-2i\omega_1 t}\left( \frac{1}{2}-\frac{1}{2}\cos({\frac{2\pi x}{L})} \right)-\beta^2 \omega_2 e^{-2i\omega_2 t}\left( \frac{1}{2}-\frac{1}{2}\cos({\frac{4\pi x}{L})} \right)dx$$

and doing the integral

$$\frac{-2\hbar}{L (\alpha^2 + \beta^2)}\left[ \frac{\alpha^2 \omega_1 e^{-2i\omega_1 t}x}{2}-\frac{\alpha^2 \omega_1 e^{-2i\omega_1 t}L}{2\pi}\sin{\frac{2\pi x}{L}}- \frac{\beta^2 \omega_2 e^{-2i\omega_2 t}}{2}-\frac{\beta^2 \omega_2 e^{-2i\omega_2 t}L}{4 \pi} \sin{\frac{4\pi x}{L}} \right]_0^L$$ $$=\frac{-\hbar \alpha^2 \omega_1 e^{-2i\omega_1 t}}{(\alpha^2 + \beta^2)}+ \frac{\hbar \beta^2 \omega_2 e^{-2i\omega_2 t}}{L(\alpha^2 + \beta^2)} $$

THis is all very well, but it still looks like it depends on time, unless it's because of one of a couple of things: 1. The difference between the exponentials is a phase difference, so they might be equivalent, or since these are stationary states we're talking about we can treat them as constants. But I wanted to make sure there wasn't some mathematical point I wasn't missing. I feel like I am almost there but not quite.

I also suspect I didn't need to do the full integration but I am not so expert with Dirac notation. I also kind of wanted to see what was "under the hood" so to speak.

and sorry for the long post.

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2 Answers 2

up vote 2 down vote accepted

The expectation value is actually $$ \langle\hat H\rangle = \int \psi^* \hat H\psi{\text d}x $$ where $\psi^*$ denotes the complex conjugate. This will only affect the exponential here, and $\exp(ix)^*\exp(ix) = \exp(-ix)\exp(ix) = \exp(ix-ix) = 1$. A tiny bit more algebra and you're done.

I think you also dropped a factor of $L$ in the denominator of one of your terms. Also, I don't understand where the factor of $2$ comes in in the top. Shouldn't the time derivative be $$ \frac{\partial\psi}{\partial t} = -\frac i{\alpha^2 + \beta^2} \left(\alpha\omega_1\phi_1 e^{-i\omega_1 t} + \alpha\omega_2\phi_2 e^{-i\omega_2 t}\right) $$ You also dropped factors of $\alpha$ and $\beta$, and introduced a negative sign when simplifying the integrand.

The following bits of intuition ought to help:

  • $\alpha$ and $\beta$ are amplitudes. They only make sense in the "quantum" world. When you take the expectation value, you're looking at a "classical" value, and they won't appear except as probabilities, when they're squared. (I'm think they never appear except when squared, but I'm not certain.)
  • The hamiltonian is energy. It's positive, unless you have a good excuse for it not being positive (like if you're examining tunnelling.)
  • The hamiltonian is energy. Check your units. The second term doesn't have the right ones.
  • Physics is simple. Complexity generally doesn't appear out of nowhere. If an answer you get is complicated, you should see one or more clear reasons why it should be. In this case, the answer you got was asymmetric (one term has a factor of $\frac 2 L$, the other didn't). Looking at the original problem, it's not clear why this should be the case, so the $\frac 2 L$ naturally falls under suspicion.
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I forgot that since the exponentials don't contain real numbers the complex conjugate isn't just the function itself again. So that means way back in my original integral I really should have had: $(\alpha \phi_1 e^{-i\omega_1 t})^*$ (I wrote it that way to show it's the conjugte) in each term, and when you multiply that out you get what you have described, yes? –  Jesse Apr 22 at 18:48
AN thanks a lot. I feel like i almost have the math but it's always like it's just out of reach. :-) I guess I am just a dullard :-) –  Jesse Apr 22 at 18:50
@Jesse, no problem, we all make these mistakes. (Or at least I do, endlessly.) I've added a few other possible corrections. Shouldn't the hamiltonian also be positive? I feel like the sign got flipped. I think the final answer ought to be just $\hbar \alpha^2 \omega_1 + \hbar \beta^2 \omega_2$ (with some normalization junk, probably $\alpha^2 + \beta^2$ in the denominator. That has the right units, at least. –  Scott Lawrence Apr 22 at 18:52
As to your notes, yeah i seem to have an extra 2 in there for some reason. I think it was because I wanted to make a problem more general. –  Jesse Apr 22 at 18:58
I added some notes on intuition for you, just in case it's interesting. (I may have a somewhat idiosyncratic way of thinking about physics, especially in re my last bullet point.) –  Scott Lawrence Apr 22 at 19:02

In addition to bytbox's answer, here is an explanation using matrix notation. The state $\psi$ in the $\{\phi_1,\phi_2\}$ basis is given by the vector $$\boldsymbol{\psi}=\begin{pmatrix}\alpha\exp\left(\frac{E_1t}{i\hbar}\right)\\\beta\exp\left(\frac{E_2t}{i\hbar}\right)\end{pmatrix}$$ where WLOG $|\alpha|^2+|\beta|^2=1$. In the $\{\phi_1,\phi_2\}$ basis $H$ admits the matrix representation $$\mathbf{H}=\begin{pmatrix} E_1&0\\ 0&E_2 \end{pmatrix} $$ and thus the problem becomes a simple matrix multiplication, $$\frac{d}{dt}\langle E\rangle=\frac{d}{dt}\left(\boldsymbol{\psi}^\dagger\mathbf{H}\boldsymbol{\psi}\right)=\frac{d}{dt}\left(|\alpha|^2E_1+|\beta|^2E_2\right)=0.$$

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Well that DOES look simpler. :-) Much appreciated.-- BTW the dagger mark is the same as the *? Means complex conjugate? –  Jesse Apr 22 at 22:15
Yep, the dagger mark is the same. (Rule-of-thumb: if there's a weird smudge above the number, 50% odds it means complex conjugate :P ) –  Scott Lawrence Apr 22 at 22:21
@Jesse: Almost, the $\dagger$ means adjoint, which is conjugate transpose. The conjugate transpose $\boldsymbol{\psi}^\dagger$ of a column vector $\boldsymbol{\psi}$ is a row vector whose entries are the conjugates of the original column vector. –  DumpsterDoofus Apr 22 at 23:35
@Jesse: It looks a bit simpler because continuous integration over real space (in the "position basis") is replaced by $2\times 2$ discrete matrix multiplication when you represent the problem in the "eigenstate basis $\{\phi_1,\phi_2\}$". It can be a helpful trick for thinking about things, and the matrices and vectors can be played with using computer algebra systems like Mathematica, which can sometimes eliminate the need to do complex algebraic manipulations by hand. –  DumpsterDoofus Apr 22 at 23:41

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