I have the correct answer except with a negative sign.

The wave function is given as, $$\Phi=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]$$

By squaring the momentum quantity, I found the expectation value of momentum squared to be $\langle-\hbar^2\frac {\partial^2}{\partial x^2}\rangle$.

I then computed the second derivative of $\Phi$ and found it to be $$\frac{\partial^2\psi}{\partial x^2}=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]\cdot\left[4\left(\frac {am}\hbar\right)^2x^2-\left(\frac {am}\hbar\right)\right].$$

The expectation value can therefore be written as $$-\hbar^2 4\left(\frac {am}\hbar\right)^2(\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx -\frac {am}\hbar\int \Phi^*\Phi \mathop{}\!\mathrm dx)$$

$\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx$ is just the expectation value for $x^2$, and the other integral is just 1 (since the wave function is normalized).

I previously found the expectation value $x^2$ to be $\frac \hbar{4ma}$.

The expectation value of momentum squared should then simplify to $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\cdot(\frac \hbar{4ma}-\frac {am}\hbar)=-\hbar am +4a^3m^3$$ The given answer is $$\hbar am$$.

I have the correct answer except with a negative sign.

The wave function is given as, $$\Phi=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]$$

By squaring the momentum quantity, I found the expectation value of momentum squared to be $\langle-\hbar^2\frac {\partial^2}{\partial x^2}\rangle$.

I then computed the second derivative of $\Phi$ and found it to be $$\frac{\partial^2\psi}{\partial x^2}=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]\cdot\left[4\left(\frac {am}\hbar\right)^2x^2-\left(\frac {am}\hbar\right)\right].$$

The expectation value can therefore be written as $$-\hbar^2 4\left(\frac {am}\hbar\right)^2(\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx -\frac {am}\hbar\int \Phi^*\Phi \mathop{}\!\mathrm dx)$$

$\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx$ is just the expectation value for $x^2$, and the other integral is just 1 (since the wave function is normalized).

I previously found the expectation value $x^2$ to be $\frac \hbar{4ma}$.

The expectation value of momentum squared should then simplify to $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\cdot(\frac \hbar{4ma}-\frac {am}\hbar)=-\hbar am +4a^3m^3$$ The given answer is $$\hbar am.$$

I have the correct answer except with a negative sign.

The wave function is given as, $$\Phi=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]$$

By squaring the momentum quantity, I found the expectation value of momentum squared to be $\langle-\hbar^2\frac {\partial^2}{\partial x^2}\rangle$.

I then computed the second derivative of $\Phi$ and found it to be $$\frac{\partial^2\psi}{\partial x^2}=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]\cdot\left[4\left(\frac {am}\hbar\right)^2x^2-\left(\frac {am}\hbar\right)\right].$$

The expectation value can therefore be written as $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx -\frac {am}\hbar\int \Phi^*(x)\Phi \mathop{}\!\mathrm dx)$$

$\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx$ is just the expectation value for $x^2$, and the other integral for $x$.

I previously found the expectation value of $x$ to be zero and the expectation of $x^2$ to be $\frac \hbar{4ma}$.

The expectation value of momentum squared should then simplify to $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\cdot\frac \hbar{4ma}=-\hbar am$$ This is exactly the negative of the given value for the answer (along with a negative expected value making no sense).

I have the correct answer except with a negative sign.

The wave function is given as, $$\Phi=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]$$

By squaring the momentum quantity, I found the expectation value of momentum squared to be $\langle-\hbar^2\frac {\partial^2}{\partial x^2}\rangle$.

I then computed the second derivative of $\Phi$ and found it to be $$\frac{\partial^2\psi}{\partial x^2}=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]\cdot\left[4\left(\frac {am}\hbar\right)^2x^2-\left(\frac {am}\hbar\right)\right].$$

The expectation value can therefore be written as $$-\hbar^2 4\left(\frac {am}\hbar\right)^2(\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx -\frac {am}\hbar\int \Phi^*\Phi \mathop{}\!\mathrm dx)$$

$\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx$ is just the expectation value for $x^2$, and the other integral is just 1 (since the wave function is normalized).

I previously found the expectation value $x^2$ to be $\frac \hbar{4ma}$.

The expectation value of momentum squared should then simplify to $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\cdot(\frac \hbar{4ma}-\frac {am}\hbar)=-\hbar am +4a^3m^3$$ The given answer is $$\hbar am$$.

I have the correct answer except with a negative sign.

The wave function is given as, $$\Phi=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]$$

By squaring the momentum quantity, I found the expectation value of momentum squared to be $\langle-\hbar^2\frac {\partial^2}{\partial x^2}\rangle$.

I then computed the second derivative of $\Phi$ and found it to be $$\frac{\partial^2\psi}{\partial x^2}=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]\cdot\left[4\left(\frac {am}\hbar\right)^2x^2-\left(\frac {am}\hbar\right)x)\right].$$

The expectation value can therefore be written as $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx -\frac {am}\hbar\int \Phi^*(x)\Phi \mathop{}\!\mathrm dx)$$

$\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx$ is just the expectation value for $x^2$, and the other integral for $x$.

I previously found the expectation value of $x$ to be zero and the expectation of $x^2$ to be $\frac \hbar{4ma}$.

The expectation value of momentum squared should then simplify to $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\cdot\frac \hbar{4ma}=-\hbar am$$ This is exactly the negative of the given value for the answer (along with a negative expected value making no sense).

I have the correct answer except with a negative sign.

The wave function is given as, $$\Phi=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]$$

By squaring the momentum quantity, I found the expectation value of momentum squared to be $\langle-\hbar^2\frac {\partial^2}{\partial x^2}\rangle$.

I then computed the second derivative of $\Phi$ and found it to be $$\frac{\partial^2\psi}{\partial x^2}=A\exp\left[-a\left(\frac{mx^2}{\hbar} + it\right)\right]\cdot\left[4\left(\frac {am}\hbar\right)^2x^2-\left(\frac {am}\hbar\right)\right].$$

The expectation value can therefore be written as $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx -\frac {am}\hbar\int \Phi^*(x)\Phi \mathop{}\!\mathrm dx)$$

$\int \Phi^*(x^2)\Phi\mathop{}\!\mathrm dx$ is just the expectation value for $x^2$, and the other integral for $x$.

I previously found the expectation value of $x$ to be zero and the expectation of $x^2$ to be $\frac \hbar{4ma}$.

The expectation value of momentum squared should then simplify to $$-\hbar^2 4\left(\frac {am}\hbar\right)^2\cdot\frac \hbar{4ma}=-\hbar am$$ This is exactly the negative of the given value for the answer (along with a negative expected value making no sense).