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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?

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    $\begingroup$ Position is an operator, so it shouldn't be a parameter inside a ket. $\endgroup$
    – Sandejo
    Commented Oct 5, 2021 at 1:56

1 Answer 1

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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

  1. The real-space wavefunction $\psi(x) = \langle x | \Psi \rangle$ appears instead of the momentum-space wavefunction.
  2. A differential operator appears at the step $\langle x | \frac{\hat{p}^2}{2m} | x' \rangle$. The easiest thing is just to use the rule that you can replace this combination with $-\frac{\hbar^2}{2m}\delta(x-x') \nabla_x^2$; then you will end up with your second derivation.
  3. As an alternative to 2, you can also proceed by inserting a complete set of momentum states, evaluating $\hat{p}$ in the momentum basis, then using $\langle x | p \rangle = \frac{1}{\sqrt{2\pi}}e^{i p x/\hbar}$ and $p e^{i p x/\hbar} = -i \hbar \nabla_x e^{i p x/\hbar}$ to convert the factors of $p$ into gradients.

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.

The key pieces of information used are

  • $\langle p | x \rangle = e^{i p x/\hbar}/\sqrt{2}$
  • \hat{p}|p\rangle = p | p\rangle
  • $p e^{i p x/\hbar} = -i \hbar \nabla_x e^{i p x/\hbar}$ (to see this just take the derivative on the RHS and you'll see it equals the LHS)
  • $p^2 e^{i p x/\hbar} = - \hbar^2 \nabla_x^2 e^{i p x/\hbar}$ (just the above equation, twice)
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  • $\begingroup$ Thank you so much for providing clarity. It is especially shameful, that an extremely important exam in India, gave the following question, and all the options were incorrect. Apart from that, could you kindly elaborate on method $3$. ? I seem to be quite lost regarding that one. $\endgroup$ Commented Oct 5, 2021 at 17:09
  • $\begingroup$ @NakshatraGangopadhay I added some comments. $\endgroup$
    – Andrew
    Commented Oct 5, 2021 at 20:04
  • $\begingroup$ Sorry to bother you after so long. I had a follow-up question to this. Suppose, instead of simple scaling, the wavefunction becomes $\psi(u)=\psi(g(x))$. Using the chain rule, I can solve this easily. However, I'm unable to understand how I should represent this in dirac notation, if I want to work with the $u$ basis directly. I know that $\psi(x)$ becomes $\psi(u)$. The operator in $x$ basis is $f(x)$. Here it becomes some $h(u)$. However, the differential $dx$ becomes $\frac{du}{g'(x)}$. I'm failing to see where this factor is coming from in the dirac notation. $\endgroup$ Commented Oct 25, 2021 at 6:46
  • $\begingroup$ $\int \psi^*(x)f(x)\psi(x) dx= \int \psi^*(g(x))f(x)\psi(g(x))dx = \int \psi^*(u)h(u)\psi(u)\frac{du}{g'(g^{-1}(u)}$. How is this chan rule, represented in Dirac notation. This extra factor of $frac{1}{g'(g^{-1}(u)}$, where is this in the dirac notation, when I enter the identity operator. $\endgroup$ Commented Oct 25, 2021 at 6:50
  • $\begingroup$ $I = \int |x\rangle\langle x| dx = \int |g(x)\rangle \langle g(x)| dg(x)$. Is this correct ? If so, then I can freely input this, in the Bra-ket notation, and proceed just like I would proceed in $x$ or $p$ basis. However, there is this extra factor under the differential arising out of the chain rule, that I can't get rid of. $\endgroup$ Commented Oct 25, 2021 at 6:54

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