Lagrangian density for a single-spin 0-real-bosonic field ($\phi$) is given by, $$\mathcal{L}=-\frac{1}{2}\partial_\mu \phi \partial^\mu \phi-\frac{m^2}{2}\phi^2$$ Now if we formulate the Euler Lagrange equation for the above Lagrangian density, we will get, $$\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial \phi_{,\mu}}\right)=\frac{\partial \mathcal{L}}{\partial \phi}$$ $$\Rightarrow -\frac{1}{2}\left[\partial_\mu\bigg(\frac{\partial}{\partial\phi_{,\mu}}\bigg)\right]\underbrace{(\partial_\nu\phi g^{\nu\alpha}\partial_\alpha\phi)}_{\text{New dummy indices, $\nu$ and $\alpha$}}=-m^2\phi$$ $$\Rightarrow -\frac{1}{2}\partial_\mu g^{\nu \alpha}(\delta^{\mu}_\nu\partial_\alpha\phi+\delta^\mu_\alpha \partial_\nu\phi)=-m^2\phi$$ $$\Rightarrow -\frac{1}{2}\partial_\mu(\partial^\mu\phi+\partial^\mu\phi)=-m^2\phi$$ $$\Rightarrow \partial_\mu\partial^\mu\phi-m^2\phi=0$$ So, in the end, we get the Klein Gordon Equation, which is a relativistic version of the Schrodinger Equation. Now my question is Euler Lagrange Equation is a classical dynamical equation, which gives a particular type of path in which the corresponding action is an extremum. But according to Feynman Path Integral formalism, in any quantum theory, all possible types of path has some finite probability (Unlike the classical case, where only a particular path, for which action will be extremum, will have a sharp probability, and the probability of other paths tend to 0).$$\\$$ So, my question is,  how the classical Euler Lagrange equation can formulate a quantum dynamical equation? How Euler Lagrange Equation is valid in Quantum Theory  $$$$  N.B.- Same thing I also have noticed in Dirac Field also.
Why is it that the classical path gives the dominant contribution in the quantum mechanical path integral? How do we understand this?
Cards changing mana color What cards across all sets allow the utilization of mana, either colorless or otherwise, as colored mana of another color?  One early example with an unlimited  1:1 ratio:     Sunglasses of Urza   Continuous Artifact: White mana in your mana pool can also be used as red mana.   What other cards have similar abilities? Shape of confidence interval for predicted values in linear regression PostgreSQL check bloating of a table How do I join several objects into one? Caps lock indicator not synced I have a laptop which I hook up to an external display when I'm at home. I also connected an external keyboard and mouse which makes it more comfortable for me.  I have noticed that pressing caps lock on the external keyboard does not light up the capslock indicator of the laptop keyboard (and vice versa).  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However, the answer key provided is $-1$ for the first question and also $-1$ for the second one. Please help. I had given an cipher text (of JSON string), an IV and a key. I tried decrypting using 's method:      decryptWithManagedIV(algorithmName, privateKey, IVAndCipherText)     //Sample code   Blob key = Blob.valueOf(decryptionKey);   Blob encryptedDatafor = EncodingUtil.base64Decode(bodystring);   Blob decryptedData = Crypto.decryptwithManagedIV('AES128', key, encryptedDatafor);    cipher = decryptedData.toString();   System.debug('Encrypted BodyString: ' +cipher);   and I was able to decrypt the cipher successfully. But when I use the provided IV with the following method:      decrypt(algorithmName, privateKey, initializationVector, cipherText)    //Sample code  Blob exampleIV = Blob.valueOf('HSDlUkhrmfEo7SgM'); //provided IV   Blob key = Blob.valueOf(decryptionKey);  Blob encryptedDatafor = EncodingUtil.base64Decode(bodystring);  Blob decryptedDataIV = Crypto.decrypt('AES128', key, exampleIV, encryptedDatafor);   cipher = EncodingUtil.base64Encode(decryptedDataIV );  System.debug('Encrypted BodyString: ' +cipher);   I get some string which looks like, it is still in the encrypted form (different from the encrypted string I have passed).   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