Suppose, I have a wavefunction given by $\psi(x,t)$. This wavefunction, over time, becomes $\psi(\alpha x,t)$. I've been asked to compute the final kinetic energy of this new wavefunction, in terms of the initial kinetic energy.
We know, $$\langle T_i\rangle=\langle \psi(x)|(-\frac{\hbar}{2m} \nabla_x^2)|\psi(x)\rangle$$
This is the initial kinetic energy, in Bra-Ket notation. We can write the final kinetic energy as :
$$\langle T_f\rangle=\langle \psi(\alpha x)|(-\frac{\hbar}{2m} \nabla_x^2)|\psi(\alpha x)\rangle$$
However, changing variable to $u$ such that $u=\alpha x$, we can see :
$$\frac{\partial}{\partial u} = \frac{\partial}{\partial x}\frac{\partial x}{\partial u} = \frac{1}{ \alpha}\frac{\partial}{\partial x}$$
Thus,
$$\alpha^2\frac{\partial^2}{\partial u^2} =\frac{\partial^2}{\partial x^2}$$
Thus, we can write kinetic energy as :
$$\langle T_f\rangle= \alpha^2\langle \psi(u)|(-\frac{\hbar}{2m} \nabla_u^2)|\psi(u)\rangle = \alpha^2\langle T_i \rangle$$
However, if I write this same thing through integration, I'm facing a problem.
$$\langle T_i \rangle = \int\psi^*(x)(-\frac{\hbar}{2m} \nabla_x^2)\psi(x)dx$$
Similarly, we have :
$$\langle T_f \rangle = \int\psi^*(u)(-\frac{\hbar}{2m} \nabla_x^2)\psi(u)dx$$
As we have seen, $$\nabla_x^2 = \alpha^2\nabla_u^2 \space\space\space\& \space\space\space dx=\frac{du}{\alpha}$$
Plugging these two values in, and noting that $u$ is just a dummy variable, we have :
$$\langle T_f \rangle = \int\psi^*(u)(-\alpha^2\frac{\hbar}{2m} \nabla_u^2)\psi(u)\frac{du}{\alpha} = \alpha \langle T_i\rangle$$
Even though the two notations are equivalent, there are giving me different answers. Can someone guide me as to where I'm making a mistake, and how should I deal with problems such as these?
Best Answer
Your second derivation is ok.
In bra-ket notation, $|\psi(x)\rangle$ is meaningless. The state is $|\Psi\rangle$, an abstract Hilbert-space vector with no explicit dependence on any variables specific to a given basis. The wavefunction $\psi(x)$ is the state projected into the position basis, $\psi(x) = \langle x | \Psi \rangle$.
Furthermore, you shouldn't write the energy operator in terms of $\nabla_x$ in bra-ket notation. The Hamiltonian is $\hat{H}=\hat{p}^2/2m$, where $\hat{H}$ and $\hat{p}$ are abstract operators. You should only express them in terms of a number or function or differential operator in some basis.
So the following expressions can be parsed \begin{equation} \langle \Psi | \hat{H} | \Psi \rangle = \langle \Psi | \frac{\hat{p}^2}{2m} | \Psi \rangle \end{equation} while you should avoid writing things like $\langle \psi(x) | \hat{H} | \psi(x) \rangle$ or $\langle \Psi | \nabla_x^2 | \Psi \rangle $ or $\langle \psi(x) | \nabla_x^2 | \psi(x) \rangle$.
To express this in terms of a basis, you insert a complete set of states using the resolution of the identity \begin{equation} \hat{\mathbf{1}} = \int d x | x \rangle \langle x | = \int \frac{dp}{2\pi} | p \rangle \langle p | \end{equation} where $\hat{\mathbf{1}}$ is the identity operator, and I've written the identity operator in two different bases (position and momentum) to emphasize that you can choose to do the calculation in any basis. The $2\pi$ is conventional (but needs to appear somewhere to make the Fourier transforms work out).
To illustrate this, let's first evaluate the expression in the momentum basis. Then \begin{eqnarray} \langle \Psi | \hat{H} | \Psi \rangle &=& \langle \Psi | \frac{\hat{p}^2}{2m}| \Psi \rangle \\ &=& \langle \Psi | \hat{\mathbf{1}} \frac{\hat{p}^2}{2m} \hat{\mathbf{1}} | \Psi \rangle \\ &=& \int \frac{d p}{2\pi} \int \frac{d p'}{2\pi} \langle \Psi | p\rangle \langle p | \frac{\hat{p}^2}{2m} | p' \rangle \langle p' | \Psi \rangle \\ &=& \int dp \int dp' \tilde{\psi}(p)^\star \left( \frac{p^2}{2m} 2\pi \delta(p-p') \right) \tilde\psi(p') \\ &=& \frac{1}{2m} \int \frac{dp}{2\pi} p^2 |\tilde{\psi}(p)|^2 \end{eqnarray} where $\tilde{\psi}(p) \equiv \langle p | \Psi \rangle$ is the state in the momentum representation, or the momentum-space wavefunction. Note that at no point in this derivation did any differential operator appear. Instead, when it came time to evaluate $\langle p | \frac{\hat{p}^2}{2m} | p' \rangle$, we only had to use $\hat{p}^2 | p \rangle = p^2 | p \rangle$ and $\langle p | p'\rangle = 2\pi \delta(p-p')$.
We can follow the exact same logic in the position representation, by replacing $\hat{\mathbf{1}}= \int dx | x \rangle \langle x | $. I'll leave the details for you (feel free to ask follow up questions). The main differences with respect to the momentum-space derivation are that
Here are some more equations to flesh out method 3. I'm not going to do the whole derivation, but just focus on the tricky expectation value $\langle x | \frac{\hat{p}^2}{2m} | x' \rangle$ that appears when you replace $\hat{\mathbf{1}}$ with $\int dx |x\rangle \langle x |$ \begin{eqnarray} \langle x | \frac{\hat{p}^2}{2m} | x' \rangle &=& \int \frac{dp}{2\pi} \int \frac{dp'}{2\pi} \langle x | p \rangle \langle p | \frac{\hat{p}^2}{2m} | p' \rangle \langle p' | x' \rangle \\ &=& \int \frac{dp}{2\pi} \int \frac{dp'}{2\pi} \frac{e^{ipx/\hbar}}{\sqrt{2\pi}} \langle p | \frac{\hat{p}^2}{2m} | p' \rangle \frac{e^{-ip'x'/\hbar}}{\sqrt{2\pi}} \\ &=& \int \frac{dp}{2\pi} \int \frac{dp'}{2\pi} \frac{e^{i(px - p'x')/\hbar}}{2\pi} \left(2\pi \delta(p-p')\frac{p^2}{2m}\right) \\ &=& \frac{1}{4\pi m} \int \frac{dp}{2\pi} p^2 e^{i p (x-x')/\hbar} \\ &=& \frac{1}{4\pi m} \int \frac{dp}{2\pi} \left(-\hbar^2 \nabla_x^2 e^{i p(x-x')/\hbar}\right) \end{eqnarray} In practice, you'll be integrating this expression over $x$ and $x'$, times wavefunctions $\psi(x)$ and $\psi(x')$, so the next step is to integrate the $\nabla_x^2$ by parts so you put it on $\psi(x)$. Then you can do the integral over $p$, giving you a delta function $\delta(x-x')$, which kills the $x'$ integral. Then you are back to your second derivation.
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