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1. Let us assume the integral form $$ \psi(x)~=~ \frac{2m}{\hbar^2} \int^{x}\mathrm{d}y \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z) \tag{1}$$ of the time independent 1D Schrödinger equation (TISE), where the potential $$V(x)~=~V_0\delta(x)\tag{2}$$ is a Dirac delta potential.

2. Let us assume that the wavefunction $\psi\in {\cal L}_{\rm loc}^1(\mathbb{R})$ is a locally integrable function, so that the integral $\int^{y}\mathrm{d}z\ E\psi(z)$ in eq. (1) is well-defined.

3. Note in particular that we do not assume that $\psi$ is continuous, cf. OP's title question. The following crucial question arises: How should we define the integral $$\int^{y}\mathrm{d}z\ V(z)\psi(z)\tag{3}$$ in eq. (1)?

   • Should we define (3) to be equal to $$V_0\theta(y)\psi(0)+C,\tag{3'}$$ where $C$ is an integration constant?

   • Or if the left and right limits $\psi(0^-)$ and $\psi(0^+)$ exist, should we define (2) to be equal to $$V_0\theta(y)\frac{\psi(0^-)+\psi(0^+)}{2}+C~?\tag{3''}$$

4. Whatever definition we propose for the Dirac delta distribution, it seems likely that (3) will become a locally integrable function of $y$.

   A mathematical bootstrap argument involving the TISE (1) then shows that $\psi\in C(\mathbb{R})$ is continuous, cf. OP's title question. See also e.g. my related Phys.SE answer here.

Well, let's carefully analyze the system:

1. Let us assume the integral form $$ \psi(x)~=~ \frac{2m}{\hbar^2} \int^{x}\mathrm{d}y \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z) \tag{1}$$ of the time independent 1D Schrödinger equation (TISE), where the potential $$V(x)~=~V_0\delta(x)\tag{2}$$ is a Dirac delta potential.

2. Let us assume that the wavefunction $\psi\in {\cal L}_{\rm loc}^1(\mathbb{R})$ is a locally integrable function, so that the integral $\int^{y}\mathrm{d}z\ E\psi(z)$ in eq. (1) is well-defined.

3. Note in particular that we do not assume that $\psi$ is continuous, cf. OP's title question. The following crucial question arises: How should we define the integral $$\int^{y}\mathrm{d}z\ V(z)\psi(z)\tag{3}$$ in eq. (1)?

   • Should we define (3) to be equal to $$V_0\theta(y)\psi(0)+C,\tag{3'}$$ where $C$ is an integration constant?

   • Or if the left and right limits $\psi(0^-)$ and $\psi(0^+)$ exist, should we define (3) to be equal to $$V_0\theta(y)\frac{\psi(0^-)+\psi(0^+)}{2}+C~?\tag{3''}$$

4. Whatever definition we propose for the Dirac delta distribution, it seems likely that (3) will become a locally integrable function of $y$.

    

5. A mathematical bootstrap argument involving the TISE (1) then shows that $\psi\in C(\mathbb{R})$ is continuous, cf. OP's title question. This conclusion is largely independent of how we defined (3).

6. See also e.g. my related Phys.SE answer here.

1. Let us assume the integral form $$ \psi(x)~=~ \frac{2m}{\hbar^2} \int^{x}\mathrm{d}y \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z) \tag{1}$$ of the time independent 1D Schrödinger equation (TISE), where the potential $$V(x)~=~V_0\delta(x)\tag{2}$$ is a Dirac delta potential.

2. Let us assume that the wavefunction $\psi\in {\cal L}_{\rm loc}^1(\mathbb{R})$ is a locally integrable function, so that the integral $\int^{y}\mathrm{d}z\ E\psi(z)$ in eq. (1) is well-defined.

3. Note in particular that we do not assume that $\psi$ is continuous, cf. OP's title question. The following crucial question arises: How should we define the integral $$\int^{y}\mathrm{d}z\ V(z)\psi(z)\tag{3}$$ in eq. (1)?

   • Should we define (3) to be equal to $$V_0\theta(y)\psi(0)+C,\tag{3'}$$ where $C$ is an integration constant?

   • Or if the left and right limits $\psi(0^-)$ and $\psi(0^+)$ exist, should we define (2) to be equal to $$V_0\theta(y)\frac{\psi(0^-)+\psi(0^+)}{2}+C~?\tag{3''}$$

4. Whatever definition we propose for the Dirac delta distribution, it seems likely that (3) will become a locally integrable function of $y$.

   A mathematical bootstrap argument involving the TISE (1) then shows that $\psi\in C(\mathbb{R})$ is continuous, cf. OP's title question. See also e.g. my related Phys.SE answer here.

1. Let us assume the integral form $$ \psi(x)~=~ \frac{2m}{\hbar^2} \int^{x}\mathrm{d}y \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z) \tag{1}$$ of the time independent 1D Schrödinger equation (TISE), where the potential $$V(x)~=~V_0\delta(x)\tag{2}$$ is a Dirac delta potential.

2. Let us assume that the wavefunction $\psi\in {\cal L}_{\rm loc}^1(\mathbb{R})$ is a locally integrable function, so that the integral $\int^{y}\mathrm{d}z\ E\psi(z)$ in eq. (1) is well-defined.

3. Note in particular that we do not assume that $\psi$ is continuous, cf. OP's title question. The following crucial question arises: How should we define the integral $$\int^{y}\mathrm{d}z\ V(z)\psi(z)\tag{3}$$ in eq. (1)?

   • Should we define (3) to be equal to $$V_0\theta(y)\psi(0)+C,\tag{3'}$$ where $C$ is an integration constant?

   • Or if the left and right limits $\psi(0^-)$ and $\psi(0^+)$ exist, should we define (2) to be equal to $$V_0\theta(y)\frac{\psi(0^-)+\psi(0^+)}{2}+C~?\tag{3''}$$

4. Whatever definition we propose for the Dirac delta distribution, it seems likely that (3) will become a locally integrable function of $y$.

   A mathematical bootstrap argument involving the TISE (1) then shows that $\psi\in C(\mathbb{R})$ is continuous, cf. OP's title question. See also e.g. my related Phys.SE answer here.

1. Let us assume the integral form $$ \psi(x)~=~ \frac{2m}{\hbar^2} \int_{-\infty}^{x}\mathrm{d}y \int_{-\infty}^{y}\mathrm{d}z\ (V(z)-E)\psi(z) \tag{1}$$ of the time independent 1D Schrödinger equation (TISE), where the potential $V(x)=V_0\delta(x)$ is the Dirac delta potential.

2. Let us assume that the wavefunction $\psi\in {\cal L}^1(\mathbb{R})$ is an integrable function, so that the $\int_{-\infty}^{y}\mathrm{d}z\ E\psi(z)$ term in eq. (1) is well-defined.

3. How should we define the $$\int_{-\infty}^{y}\mathrm{d}z\ V(z)\psi(z)\tag{2}$$ term in eq. (1)?

   • Should we define (2) to be equal to $$V_0\theta(y)\psi(0)~?\tag{2'}$$

   • Or if the left and right limits $\psi(0^-)$ and $\psi(0^+)$ exist, should we define (2) to be equal to $$V_0\theta(y)\frac{\psi(0^-)+\psi(0^+)}{2}~?\tag{2''}$$

4. Whatever definition we propose for the Dirac delta distribution, it seems likely that (2) will become a locally integrable function of $y$.

   A mathematical bootstrap argument involving the TISE (1) then shows that $\psi\in C(\mathbb{R})$ is continuous. See also e.g. my related Phys.SE answer here.

1. Let us assume the integral form $$ \psi(x)~=~ \frac{2m}{\hbar^2} \int^{x}\mathrm{d}y \int^{y}\mathrm{d}z\ (V(z)-E)\psi(z) \tag{1}$$ of the time independent 1D Schrödinger equation (TISE), where the potential $$V(x)~=~V_0\delta(x)\tag{2}$$ is a Dirac delta potential.

2. Let us assume that the wavefunction $\psi\in {\cal L}_{\rm loc}^1(\mathbb{R})$ is a locally integrable function, so that the integral $\int^{y}\mathrm{d}z\ E\psi(z)$ in eq. (1) is well-defined.

3. Note in particular that we do not assume that $\psi$ is continuous, cf. OP's title question. The following crucial question arises: How should we define the integral $$\int^{y}\mathrm{d}z\ V(z)\psi(z)\tag{3}$$ in eq. (1)?

   • Should we define (3) to be equal to $$V_0\theta(y)\psi(0)+C,\tag{3'}$$ where $C$ is an integration constant?

   • Or if the left and right limits $\psi(0^-)$ and $\psi(0^+)$ exist, should we define (2) to be equal to $$V_0\theta(y)\frac{\psi(0^-)+\psi(0^+)}{2}+C~?\tag{3''}$$

4. Whatever definition we propose for the Dirac delta distribution, it seems likely that (3) will become a locally integrable function of $y$.

   A mathematical bootstrap argument involving the TISE (1) then shows that $\psi\in C(\mathbb{R})$ is continuous, cf. OP's title question. See also e.g. my related Phys.SE answer here.